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Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B Direction Cosines and Direction Ratios Important Questions

Question 1.
If P (2, 3, -6), Q (3, -4, 5) are two points, find the d.c.s of \(\overrightarrow{\mathrm{OP}}\), \(\overrightarrow{\mathrm{QO}}\) and \(\overrightarrow{\mathrm{PQ}}\) where O is the origin.
Solution:

Question 2.
Find the d.c.s of a line that makes equal angles with the axes.
Solution:
Suppose the line makes angle a with the axes d.cs of the line are (cos α, cos α, cos α)
But l² + m² + n² = 1
cos² α + cos² α + cos² α = 1
3 cos² α = 1 ⇒ cos α = \(\frac{1}{3}\)
cos α = ± \(\frac{1}{\sqrt{3}}\)
d.cs of the required line are
(± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\))
Hence there are 8 such directions which reduce to 4 lines.

Question 3.
If the d.cs of a line are (\(\frac{1}{c}\), \(\frac{1}{c}\), \(\frac{1}{c}\)) find c.
Solution:
We know that l² + m² + n² = 1
\(\frac{1}{c^{2}}\) + \(\frac{1}{c^{2}}\) + \(\frac{1}{c^{2}}\) = 1, \(\frac{3}{c^{2}}\) = 1 ⇒ c² = 3
c = ± \(\sqrt{3}\)

Question 4.
Find the direction cosines of two lines which are connected by the relations l + m + n = 0 and mn – 2nl – 2lm = 0. [Mar 11]
Solution:
Given l + m + n = 0 ………………… (1)
mn – 2nl – 2lm = 0 ………………….. (2)
From (1)l = -(m + n)
Substituting in (2)
mn ± 2n (m + n) + 2m (m + n) = 0
mn + 2mn + 2n² + 2m² + 2mn = 0
2m² + 5mn + 2n² = 0
(2m + n) (m + 2n) = 0
2m = -n or m = -2n
Case (i): 2m₁ = -n₁
From l_r = -m₁ – n₁
= -m₁ + 2m₁ = m₁
\(\frac{l_{1}}{1}\) = \(\frac{m_{1}}{1}\) = \(\frac{x_{1}}{-2}\)
D.Rs of the first line are 1, 1, -2
D.Cs of this line are \(\frac{1}{\sqrt{6}}\), \(\frac{1}{\sqrt{6}}\), –\(\frac{2}{\sqrt{6}}\)
Case (ii): m₂ = -2n₂
From (l) l₂ = -m₂ – n₂ = +2n₂ – n₂ = n₂
\(\frac{l_{2}}{1}\) = \(\frac{m_{2}}{-2}\) = \(\frac{n_{2}}{1}\)
d.cs of the second line are 1, -2, 1
d.cs of this line are \(\frac{1}{\sqrt{6}}\), \(\frac{-2}{\sqrt{6}}\), \(\frac{1}{\sqrt{6}}\)

Question 5.
A ray makes angles \(\frac{\pi}{3}\), \(\frac{\pi}{3}\) with \(\overrightarrow{\mathrm{OX}}\) and \(\overrightarrow{\mathrm{OY}}\) respectively. Find the angle made by it with \(\overrightarrow{\mathrm{OZ}}\).
Solution:
We know that l² + m² + n² = 1
cos²α + cos²β + cos²γ = 1
i.e., cos² \(\frac{\pi}{3}\) + cos² \(\frac{\pi}{3}\) + cos² γ = 1
\(\frac{1}{4}\) + \(\frac{1}{4}\) + cos² γ = 1
cos² γ = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
cos γ = ± \(\frac{1}{\sqrt{2}}\)
γ = cos^-1 (± \(\frac{1}{\sqrt{2}}\))
= \(\frac{\pi}{4}\) or \(\frac{3\pi}{4}\).

Question 6.
Find the d.r.s and d.c.s of the line joining the points (4, -7, 3), (6, -5, 2).
Solution:
d.c.s of the lines are (6 – 4, -5 + 7, 2 – 3)
= (2, 2, -1)
Dividing with \(\sqrt{4+4+1}\) = 3
d.cs of the line are ± (\(\frac{2}{3}\), \(\frac{2}{3}\), –\(\frac{1}{3}\)

Question 7.
If the d.c.s of a line are proportional the (1, -2, 1) find its d.c.s.
Solution:
d.rs of the line are (1, -2, 1 )
Dividing with \(\sqrt{6}\)
d.cs of the line are ± (\(\frac{1}{\sqrt{6}}\), –\(\frac{2}{\sqrt{6}}\), \(\frac{1}{\sqrt{6}}\))

Question 8.
Show that the line joining the points P(0, 1, 2) and Q (3, 4, 8) is parallel to the line joining the points R (-2. \(\frac{3}{2}\). – 3) and S (\(\frac{5}{2}\), 6, 6).
Solution:
DR’s of PQ are (3 – 0, 4 – 1, 8 – 2) = (3, 3, 6)
DR’s of RS are (\(\frac{5}{2}\) + 2, 6 – \(\frac{3}{2}\), 6 + 3)
= (\(\frac{9}{2}\), \(\frac{9}{2}\), 9)
d.r’s of PQ and RS are proportional since \(\frac{2}{3}\) (\(\frac{9}{2}\), \(\frac{9}{2}\), 9) = (3, 3, 6)
∴ PQ is parallel to RS.

Question 9.
Show that the line joining the points A (2, 3, -1) and B(3, 5, -3) is perpendicular to the Line joining C(1, 2, 3) and D(3, 5, 7).
Solution:
d.rs of AB are (3 – 2, 5 – 3, -3 + 1)
= (1, 2, -2)
d.rs of CB are (3 – 1, 5 – 2, 7 – 3) = (2, 3, 4)
a₁a₂ + b₁b₂ + c₁c₂ = 1.2 + 2.3 – 2.4
= 2 + 6 – 8 = 0
∴ AB and CD are perpendicular.

Question 10.
For what value of x the line joining A(4, 1, 2) B (5, x, 0) is perpendicular to the line joining C(1, 2, 3) and D(3, 5, 7) ?
Solution:
d.r’s of AD are (1, x-1, -2)
d.rs of CD are (2, 3, 4)
AB and CD are perpendicular
a₁a₂ + b₁b₂ + c₁c₂ = 0
1.2 – 3(x – 1) + 4 (-2) = 0
2 + 3x – 3 – 8 = 0
3x = 9
⇒ x = 3

Question 11.
Show that the points A (1, 2, 3), B (4, 0, 4), C(-2, 4, 2) are collinear.
Solution:
d.rs of \(\overline{\mathrm{AB}}\) are 4 – 1, 0 – 2, 3 – 4
i.e., 3, -2, -1
d.rs of \(\overline{\mathrm{BC}}\) are -2 – 4, 4 – 0, 2 – 4
i.e., -6, 4, -2
d.rs of AB and BC are proportional and B is a common point. A, B, C are collinear.

Question 12.
A (1, 8, 4), B (0, -11, 4), C (2, -3, 1) are three points and D is the foot of the perpendicular from A to BC. Find the co-ordinates of D.
Solution:
Suppose divides BC in the ratio m = n
Co-ordinates of D are

(\(\frac{2 m-m-n}{m+n}\), \(\frac{-3 m-11 n-8 m-8 n}{m+n}\), \(\frac{m+4 n-4 m-4 n}{m+n}\))
\(\frac{m-n}{m+n}\), \(\frac{-11 m-19 n}{m+n}\), \(\frac{-3 m}{m+n}\)
d.rs of BC are (2 – 0, -3 + 11, 1 – 4) = (2, 8, -3)
AB and BC are perpendicular
\(\frac{2(m-n)}{m+n}\) + 8\(\frac{(-11 m-19 n)}{m+n}\) + \(\frac{(-3)(-3 m)}{m+n}\) = 0
2m – 2n – 88m – 152n + 9m = 0
-77m – 154n = 0
77m = -154n
m = -2n
co-ordinates of are
(\(\frac{-4 n}{-n}\), \(\frac{6 n-11 n}{-n}\), \(\frac{-2 n+4 n}{-n}\))
= (4, 5, -2)

Question 13.
Lines \(\overrightarrow{\mathrm{OA}}\), \(\overrightarrow{\mathrm{OB}}\) are drawn from O with direction cosines proportional to (1, -2, -1), (3, -2, 3). Find the direction cosines of the normal to the plane AOB.
Solution:
Suppose l, m, n are the d.cs of the normal.
It is perpendicular to every line in the plane AOB
l – 2m – n = 0
3l – 2m + 3n = 0

Question 14.
Find the angle between the diagonals of a cube.

Solution:
Let ‘O’ one of the vertices of the cube taken as origin and the three sides OA, OB, OC are taken as co-ordinate axes. Let OA = OB = OC = a the four diagonals are \(\overrightarrow{\mathrm{OF}}\), \(\overrightarrow{\mathrm{AG}}\), \(\overrightarrow{\mathrm{DE}}\) and \(\overrightarrow{\mathrm{BC}}\)
The co-ordinates of the vertices of the cube are 0(0, 0, 0), A(a, 0, 0), B(0, a, 0), C(0,0, a), F(a, a, 0), D(a, a, 0), E(a, 0, a), G(0, a, a)
D.Rs of OF are (a-0, a – 0, a – 0) = (a, a, a)
D.Rs of AG are (0 – 9, a – 0, a – 0) = (-a, a, a)
If O is the angle between the diagonals OF and AG then
cos θ = \(\frac{|a(-a)+a \cdot a+a \cdot a|}{\sqrt{a^{2}+a^{2}+a^{2}} \cdot \sqrt{a^{2}+a^{2}+a^{2}}}\)
= \(\frac{a^{2}}{3 a^{2}}\) = \(\frac{1}{3}\)
θ = Cos^-1 (\(\frac{1}{3}\))y
Similarly, the angle between any pair of diagonals can be shown to be cos^-1 (\(\frac{1}{3}\))

Question 15.
Show that the line whose d.c’s are proportional to (2, 1, 1) (4, \(\sqrt{3}\) -1, –\(\sqrt{3}\) – 1) are inclined to one another at angle \(\frac{\pi}{3}\).
Solution:
The d.cs of the given lines are (2, 1, 1), (4, \(\sqrt{3}\) – 1, –\(\sqrt{3}\) – 1) 4² + (\(\sqrt{3}\) – 1)² + (-\(\sqrt{3}\) – 1)²
= 16 + 3 + 1 – 2\(\sqrt{3}\) + 3 + 1 + 2\(\sqrt{3}\)
= 24

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Three Dimensional Coordinates Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B Three Dimensional Coordinates Important Questions

Question 1.
Show that the points A(-4, 9, 6), B(-1, 6, 6), and C(0, 7, 10) from a right-angled isosceles triangle.
Solution:
A(-4, 9, 6), B(-1, 6, 6), C(0, 7, 10) are the vertices of ∆ABC.
AB = \(\sqrt{(-4+1)^{2}+(9-6)^{2}+(6-6)^{2}}\)
= \(\sqrt{9+9}\)
= \(\sqrt{18}\)
BC = \(\sqrt{(-1-0)^{2}+(6-7)^{2}+(6-10)^{2}}\)
= \(\sqrt{1+1+16}\) = \(\sqrt{18}\)
CA = \(\sqrt{(0+4)^{2}+(7-9)^{2}+(10-6)^{2}}\)
= \(\sqrt{16+4+16}\) = \(36\)
AB = BC and AB² + BC² = CA²
ABC is a right-angled isosceles triangle.

Question 2.
Show that locus of the point whose distance from Y-axis is thrice its distance from (1, 2, -1) is 8x² + 9y² + 8z² – 18x – 36y + 18z + 54 = 0.
Solution:
Let P (x, y, z) be any point on the locus PM – distance from Y – axis = \(\sqrt{x^{2}+z^{2}}\)
A(1, 2, -1) is the given point
Given condition is PM = 3 . PA
PM² = 9PA²
x² + z² = 9[(x – 1)² + (y – 2)² + (z + 1)²]
= 9x² – 18x + 9 + 9y² – 36y + 36 + 9z² + 18z + 9
Locus of P is 8x² + 9y² + 8z² – 18x – 36y + 18z + 54 = 0
P satisfies the equation
8x² + 9y² + 8z² – 18x – 36y + 18z + 54 = 0

Question 3.
A, B, C are three points on \(\overrightarrow{o x}\), \(\overrightarrow{o y}\), \(\overrightarrow{o z}\) respectively at distances a, b, e. (a ≠ 0, b ≠ 0, c ≠ 0) from the origin 0. Find the co-ordinates of the point which is equidistant from A, B, C and O.
Solution:
A is a point on ox ⇒ co-ordinates of A are (a, 0, 0)
Similarly co-ordinates of B are (0, b, 0) and co-ordinates of C are (0, 0, c)
P(x, y, z) is the required point
PO = PA = PB = PC
PO² = PA² = PB² = PC²
PO² = PA²
x² + y² + z² = (x – a)² + y² + z²
x² – x² + a² – 2ax = 0
2ax = a² ⇒ x = \(\frac{a^{2}}{2 a}\) = \(\frac{a}{2}\)
PO² = PB² ⇒ y = b/2
PO² = PC² ⇒ z = c/2
Co-ordinates of P are (\(\frac{a}{2}\), \(\frac{b}{2}\), \(\frac{c}{2}\) )

Question 4.
Show that the points A (3, -2, 4) B(1, 1, 1) and C(-1, 4, -2) are collinear.
Solution:
A(3, -2, 4), B(1, 1, 1), C(-1, 4, -2) are the given points

A, B, C are collinear.

Question 5.
Find the ratio in which YZ-plane divides the line joining A(2, 4, 5) and B(3, 5, -4). Also find the point of intersection
Solution:
Suppose the line AB meets YZ plane in P
Let P divide AB in the ratio k : 1
Co-ordinates of P are
\(\left(\frac{3 k+2}{k+1}, \frac{5 k+4}{k+1}, \frac{-4 k+5}{k+1}\right)\)
P is a point on YZ plane
⇒ x co-ordinates of p = 0
\(\frac{3 k+2}{k+1}\) = 0 ⇒ 3k + 2 = 0
k = –\(\frac{2}{3}\)
YZ plane divides AB is the ratio -2 : 3
Substituting the value of k in co-ordinates of p, co-ordinates of p are
\(\left[0, \frac{5\left(-\frac{2}{3}\right)+4}{-\frac{2}{3}+1} \cdot \frac{(-4)\left(-\frac{2}{3}\right)+5}{-\frac{2}{3}+1}\right]\)
(0, 2, 23)

Question 6.
Show that the points A(3, -2, 4), B(1, 1, 1) and C(-1, 4, -2) are collinear.
Solution:
Suppose the point P divides AD in the ratio k : 1
Co-ordinates of P are \(\left(\frac{k+3}{k+1}, \frac{k-2}{k+1}, \frac{k+4}{k+1}\right)\)
If A, B, C are collinear, C lies on AB
P must coincide with C for same value of k
\(\frac{k+3}{k+1}\) = -1 ⇒ k + 3 = -k – 1
2k = 4 ⇒ k = -2
Substituting k = -2 in co—ordinates of P we get
\(\left(\frac{-2+3}{-2+1}, \frac{-2-2}{-2+1}, \frac{-2+4}{-2+1}\right)\)
= (-14, -2) = c
A, B, C are collinear.

Question 7.
Find the fourth vertex of the parallelogram whose consecutive vertices are (2, 4, -1), (3, 6, -1) are (4, 5, 1) [Mar 11]
Solution:

ABCD is a parallelogram
where A = (2, 4, -1), B = (3, 6, -1) C = (4, 5, 1)
Suppose D(x, y, z) is the fourth vertex
A B C D is a parallellelogram
Mid point of AC = Mid point of BD

∴ Co-ordinates of the fourth vertex are = D (3, 3, 1)

Question 8.
A(5, 4, 6), B(1, -1, 3), C(4, 3, 2) are three points. Find the co-ordinates of the point in which the bisector of ∠BAC meets the side \(\overline{B C}\).
Solution:
We know that if AB is the bisector of ∠BAC
Divides BC in the ratio AB : AC

Question 9.
If (x₁, y₁, z₁) and (x₂, y₂, z₂) are two vertices and (α, ß, γ) is the centroid of a triangle, find the third vertex of the triangle.
Solution:
Let A = (x₁, y₁, z₁) and B = (x₂, y₂, z₂) be the two vertices of the triangle ABC.

Let G = (α, β, γ) be the centroid.
If C = (x₃, y₃, z₃) is the third vertex, then we have
(\(\frac{x_{1}+x_{2}+x_{3}}{3}\), \(\frac{y_{1}+y_{2}+y_{3}}{3}\), \(\frac{z_{1}+z_{2}+z_{3}}{3}\)) = (α, β, γ)
⇒ x₁ + x₂ + x₃ = 3α; y₁ + y₂ + y₃ = 3β; z₁ + z₂ + z₃ = 3γ.
⇒ x₃ = 3α – x₁ – x₂; y₃ = 3β – y₁ – y₂; z₃ = 3γ – z₁ – z₂.
∴ The third vertex
C = (3α – x₁ – x₂, 3β – y₁ – y₂, 3γ – z₁ – z₂).

Question 10.
If D(x₁, y₁, z₁), E(x₂, y₂, z₂) and F(x₃, y₃, z₃) are the midpoints of the sides BC, CA and AB respectively of a triangle, find its vertices A, B and C.
Solution:
It is given that D is the mid point of the side BC, E is the mid point of the side CA and F is the mid point of the side AB.

∴ DEP is the triangle formed out of the mid points of the three sides.
Consider the parallelogram AEDF. Let A = (h, k, s).
Mid point of AD = Mid point of EF
⇒ (\(\frac{h+x_{1}}{2}\), \(\frac{k+y_{1}}{2}\), \(\frac{s+z_{1}}{2}\)) = (\(\frac{x_{2}+x_{3}}{2}\), \(\frac{y_{2}+y_{3}}{2}\), \(\frac{z_{2}+z_{3}}{2}\))
⇒ h = x₂ + x₃ – x₁; k = y₂ + y₃ – y₁; s = z₂ + z₃ – z₁
∴ Vertex A
= (x₂ + x₃ – x₁, y₂ + y₃ – y₁, z₂ + z₃ – z₁)
Similarly, the vertices B and C can be obtained as
B = (x₃ + x₁ – x₂, y₃ + y₁ – y₂, z₃ + z₁ – z₂)
C = (x₁ + x₂ – x₃, y₁ + y₂ – y₃, z₁ + z₂ – z₃).

Question 11.
If M(α, β, γ) is the mid point of the line segment joining the points A(x₁, y₁, z₁) and B, then find B.
Solution:

Let B(h, k, s) be the point required.
It is given that M is the mid point of AB.
∴ We have (α, β, γ) = (\(\frac{x_{1}+h}{2}\), \(\frac{y_{1}+k}{2}\), \(\frac{z_{1}+s}{2}\))
⇒ 2α = x₁ + h; 2β = y₁ + k; 2γ = z₁ + s
⇒ h = 2α – x₁; k = 2β – y₁; s = 2γ – z₁
∴ Point Bis (2α – x₁, 2β – y₁, 2γ – z₁).

Question 12.
If H, G, S and I respectively denote orthocentre, centroid, circumcentre and in-centre of a triangle formed by the points (1, 2, 3), (2, 3, 1) and (3, 1, 2), then find H, G, S, I.
Solution:

Since AB = BC = CA, ABC is equilateral triangle.
We know that orthocentre, centroid, circumcentre and incentre of equilateral triangle are the same (i.e., all the four points coincide). Now, centroid .
G = (\(\frac{1+2+3}{3}\), \(\frac{2+3+1}{3}\), \(\frac{3+1+2}{3}\))
= (2, 2 ,2)
∴ H = (2, 2, 2), S = (2, 2, 2), I = (2, 2, 2).

Question 13.
Find the incentre of the triangle formed by the points (0, 0, 0), (3, 0, 0) and (0, 4, 0).
Solution:
If a, b, care the side of the triangle ABC, where
A = (x₁, y₁, z₁), B = (x₂, y₂, z₂), C = (x₃, y₃, z₃) are the vertices, then the in-centre of the triangle is given by

Question 14.
If the point (1, 2, 3) is changed to the point (2, 3, 1) through translation of axes, find the new origin.
Solution:
Let (x, y. z) be the co-ordinates of any point P w.r.t. the co-ordinate frame Oxyz and (X, Y, Z) be the co-ordinates of P w.r.t the new frame of reference O’XYZ.

Let O’ (h, k, s) be the new origin so that
x = X + h, y = Y + k and z = Z + s.
⇒ (h, k, s) (x – X, y – Y, z – Z)
⇒ (h, k, s) = (1 – 2, 2 – 3, 3 – 1)
= (-1, -1, 2).
∴ O’ = (-1, -1, 2) is the new origin.

Question 15.
Find the ratio in which the point P(5, 4, -6) divides the line segment joining the points (3, 2, -4) and B(9, 8, -10). Also find the harmonic conjugate of P.
Solution:

∴ Q(-3, -4, 2) is the harmonic conjugate of P(5, 4, -6).

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Pair of Straight Lines Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B Pair of Straight Lines Important Questions

Question 1.
Does the equation x² + xy + y² = 0 represent a pair of lines?
Solution:
a = 1, b = 1, h = \(\frac{1}{2}\), ⇒ h² = \(\frac{1}{4}\), ab = 1
h² = ab < 0 i.e., h² < ab.
∴ The given equation does not represent a pair of lines

Question 2.
Find the nature of the triangle of formed by the lines x² – 3y² = 0 and x = 2.
Solution:
Combined equation of OA and OB is
x² – 3y² = 0
(x + \(\sqrt{3}\) y) (x – \(\sqrt{3}\) y) = 0
x + \(\sqrt{3}\) y = 0 and x – \(\sqrt{3}\) y = 0
i.e., y = \(\frac{1}{\sqrt{3}}\) x, y = –\(\frac{1}{\sqrt{3}}\) x are equally inclined to the X—axis, the inclination being 30°
∴ ∠OAB – ∠OBA = 60°
∴ The triangle is equilateral

Question 3.
Find the centroid of the triangle formed by the lines 12x² – 20xy + 7y² = 0 and 2x – 3y + 4 = 0.
Solution:
Combined equation of OA and OB is
12x² – 20xy + 7y² = 0 ………………….. (1)
Equation of AB is 2x – 3y + 4 = 0
2x = 3y – 4
Substituted in (1)
3(3y – 4)² – 1oy (3y – 4) + 7y² = 0
3(9y² + 16 – 24y) – 30y² + 40y + 7y² = 0
27y² +48 – 72y – 30y² + 40y + 7y² = 0
4y² – 32y + 48 = 0
y² – 8y + 12 = 0
(y – 2) (y – 6) = 0 ⇒ y = 2 or 6
x = \(\frac{3 y-4}{2}\)
y = 2 ⇒ x = \(\frac{6-4}{2}\) = 1
y = 6 ⇒ x = \(\frac{18-4}{2}\) = 7
7 2
∴ Vertices and O (0, 0), A (1, 2), B( 7, 6)
Centroid of OAB is
(\(\frac{0+1+7}{3}\), \(\frac{0+2+6}{3}\)) = (\(\frac{8}{3}\), \(\frac{8}{3}\))

Question 4.
Prove that the lines represented by the equations x² – 4xy + y² = 0 and x + y = 3 form an equilateral triangle.
Solution:
Since the straight line L: x + y = 3 makes 45° with the negative direction of the X-axis, none of the lines which makes 60° with the line L is vertical. If ‘m’ is the slope of one such straight line, then

\(\sqrt{3}\) = tan 60° = \(\left|\frac{m+1}{1-m}\right|\) and so, m satisfies the equation (m + 1)² = 3(m – 1)²
(or) m² – 4m + 1 = 0 …………………….. (1)
But the straight line having slope’m1 and passing through the origin is
y = mx ……………………. (2)
So the equation of the pair of lines passing through the origin and inclined at 60° with the line L is obtained by eliminating’m’ from the equations (1) and (2). Therefore the combined equation of this pair of lines is
(\(\frac{y}{x}\))² – 4(\(\frac{y}{x}\)) + 1 = 0 (i.e.,) x² – 4xy + y² = 0
Which is the same as the given pair of lines. Hence, the given triad of lines form an equilateral triangle.

Question 5.
Show that the product of the perpendicular distances from a point (α, β) to the pair of straight lines ax² + 2hxy + by² = 0 is \(\frac{\left|a \alpha^{2}+2 h \alpha \beta+b \beta^{2}\right|}{\sqrt{(a-b)^{2}+4 h^{2}}}\). [May 11, 07; Mar. 07, 04]
Solution:
Let ax² + 2hxy + by² = (l₁x + m₁y) (l₂x + m₂y)
Then the separate equations of the lines represented by the equation
ax² + 2hxy + by² = 0 are
L₁ : l₁x + m₁y = 0 and L₂ : l₁x + m₁y = 0
Also, we have l₁l₂ = a; m₁m₂ = b and
l₁m₂ + l₂m₁ = 2h
d₁ = length of the perpendicular from (α, β) to L₁ = \(\frac{\left|l_{1} \alpha+\mathrm{m}_{1} \beta\right|}{\sqrt{l_{1}^{2}+\mathrm{m}_{1}^{2}}}\)
d₂ = length of the perpendicular from (α, β) to L₂ = \(\frac{\left|l_{2} \alpha+m_{2} \beta\right|}{\sqrt{l_{2}^{2}+m_{2}^{2}}}\)
Then, the product of the lengths of the perpendiculars from (α, β) to the given pair of lines = d₁d₂
= \(\frac{\left\|\left(l_{1} \alpha+m_{1} \beta\right)\left(l_{2} \alpha+m_{2} \beta\right)\right\|}{\sqrt{\left(l_{1}^{2}+m_{1}^{2}\right)\left(l_{2}^{2}+m_{2}^{2}\right)}}\)
= \(=\frac{\left|a \alpha^{2}+2 h \alpha \beta+b \beta^{2}\right|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)

Question 6.
Let ax² + 2hxy + by² = 0 represent a pair of straight lines. Then show that the equation of the pair of straight lines.
i) Passing through (x₀, y₀) and parallel to the given pair of lines is
a(x – x₀)² + 2h(x – x₀) (y – y₀) + b(y – y₀)² = 0 and
ii) Passing through (x0, y0) and perpendicular to the given pair of lines is
b(x – x₀)² – 2h(x – x₀) (y – y₀) + a(y – y₀)² = 0
Solution:
Let ax² + 2hxy + by² = (l₁x + m₁y) (l₂x + m₂y).
Then the lines L₁ and L₂ are represented by the given equation are respectively
l₁x + m₁y = 0 and l₂x + m₂y = 0.
Also we have l₁l₂ = a, m₁m₂ = b and
l₁m₂ + l₂m₁ = 2h

(i) The straight lines passing through (x₀, y₀) and parallel to L₁ and L₂ are respectively
l₁x + m₁y = l₁x₀ + m₁y₀ (or)
l₁(x – x₀) + m₁(y – y₀) = 0 and
l₂(x -x₀) + m₂(y – y₀) = 0.
Therefore, their combined equation is
[l₁(x – x₀) + m₁(y – y₀)] [l₂(x – x₀)+ m₂(y – y₀)] = 0
(or) a(x – x₀)² + 2h(x – x₀) (y – y₀) + b(y – y₀)² = 0

(ii) The straight lines passing through (x₀, y₀) and perpendicular to the pair L₁ and L₂ are respectively.
m₁x – l₁y = m₁x₀ – l₁y₀ (or)
m₁(x – x₀) – l₁(y – y₀) = 0 and
m₂(x – x₀) – l₂(y – y₀) = 0.
Hence their combined equation is
[m₁(x – x₀) – l₁(y – y₀)] [m₂(x – x₀) – l₂(y – y₀)] = 0
(i.e.,) b(x – x₀)² – 2h(x – x₀) (y – y₀) + a(y – y₀)² = 0
[Note : The pair of lines passing through the origin and perpendicular to the pair of lines given by ax² + 2hxy + by² = 0 is bx² – 2hxy + ay² = 0].

Question 7.
Show that the area of the triangle formed by the lines ax² + 2hxy + by² = 0 and lx + my + n = 0 is \(\left|\frac{n^{2} \sqrt{h^{2}-a b}}{a m^{2}-2 h / m+b l^{2}}\right|\)
Solution:
Let \(\overleftrightarrow{\mathrm{OA}}\) and \(\overleftrightarrow{\mathrm{OB}}\) be the pair of straight lines represented by the equation
ax² + 2hxy + by² = 0 (see figure) and \(\overleftrightarrow{\mathrm{AB}}\) be the line lx + my + n = 0
Let ax² + 2hxy + by² = (l₁x + m₁y) (l₂x + m₂y) and\(\overleftrightarrow{\mathrm{OA}}\) and \(\overleftrightarrow{\mathrm{OB}}\) be the lines.
l₁x + m₁y = 0 and l₂x + m₂y = 0 respectively.

Let A = (x₁, y₁) and B = (x₂, y₂).
Then l₁x₁ + m₁y₁ = 0 and l₁x₁ + m₁y₁ + n = 0.
So, by the rule of cross-multiplication, we obtain

Question 8.
Two equal sides of an isosceles triangle are 7x – y + 3 = 0 and x + y – 3 = 0 and its third side passes through the point (1,0). Find the equation of the third side.
Solution:
Let the lines 7x – y + 3 – 0 and x + y – 3 = 0 intersect at A. If we draw lines (not passing through A) perpendicular to each of the bisectors of the angles at A, we get isosceles triangles, equal sides being along the given lines.
(∆ABF ≅ ∆AFC and ∆ADG ≅ ∆AGE)
Of them, we require those triangles whose third sides pass through (1, 0).
The equations of the bisectors of the angles between 7x – y + 3 = 0 and x + y – 3 = 0 are

⇒ 7x – y + 3 = ± 5(x + y – 3)
⇒ x – 3y + 9 = 0 and 3x + y – 3 = 0.
The third sides will be those lines perpendicular to the bisectors and intersecting at (1, 0).

The side perpendicular to x – 3y + 9 = 0 and passing through (1, 0) is 3x + y – 3 =0. The other one is (x – 1) -3(y – 0) = 0 i.e., x – 3y – 1 = 0. Therefore 3x + y- 3 = 0 and x- 3y— 1 = 0 are the required ones. [In the Figure ∆ABC and ∆ADE are isosceles with \(\overline{B C}\) and \(\overline{D E}\) as third sides].

Question 9.
Find the angle between the straight lines represented by 2x² + 5xy + 2y² – 5x – 7y + 3 = 0.
Solution:
a = 2, b = 2, h = \(\frac{5}{2}\)
cos θ = \(\frac{|a+b|}{\sqrt{(a-b)^{2}+4 h^{2}}}=\frac{|2+2|}{\sqrt{(2-2)^{2}+4 \cdot \frac{25}{4}}}\) = \(\frac{4}{5}\)
θ = cos^-1 (\(\frac{4}{5}\))

Question 10.
Find the equation of the pair of lines passing through the origin and parallel to the pair of lines 2x² + 3xy – 2y² – 5x + 5y – 3 = 0
Solution:
Equation of the given pair of lines is
2x² + 3xy – 2y² – 5x + 5y – 3 = 0
Equation of the pair of parallel lines passing through the origin is ax² + 2hxy + by² = 0 is 2x² + 3xy – 2y² = 0

Question 11.
Find the equation of the pair of lines passing through the origin and perpendicular to the pair of lines ax² + 2hxy + by² + 2gx + 2fy + c = 0
Solution:
Equation to the pair of lines parallel to the given lines and passing through the origin is ax² + 2hxy + by² = 0
Equation of the pair of lines perpendicular to the given lines and passing through the origin is bx² – 2hxy + ay² = 0

Question 12.
If x² + xy – 2y² + 4x – y + k = 0 represents a pair of straight lines, find k.
Solution:
a = 1, b = -2, c = k; f = –\(\frac{1}{2}\), g = 2, h = \(\frac{1}{2}\)
The condition is
abc + 2fgh – af² – bg² – ch² = 0
2k + 2(-\(\frac{1}{2}\)) . 2\(\frac{1}{2}\) – 1.\(\frac{1}{4}\) + 2.\(\frac{4}{a}\) – k\(\frac{1}{4}\) = 0
-8k – 4 – 1 + 8 – k = 0
9k = 27 ⇒ k = 3

Question 13.
Prove that the equation 2x² + xy – 6y² + 7y – 2 = 0 represents a pair of straight lines.
Solution:
a = 2
b = -6
c = -2
f = \(\frac{7}{2}\)
g = 0
h = \(\frac{1}{2}\)
abc + 2fgh – af² – bg² – ch²
= 2(-6) (-2) + 2.\(\frac{7}{2}\) 0\(\frac{1}{2}\) – 2(\(\frac{7}{2}\))² + 6.0 + 2.(\(\frac{1}{2}\))²
= 24 – \(\frac{49}{2}\) + \(\frac{1}{2}\) = 0
h² – ab = \(\frac{1}{4}\) + 12 > 0,
g² – ac = 0 + 4 = 4 > 0
f² – bc = \(\frac{49}{4}\) – 12 = \(\frac{1}{4}\) > 0
The given equation represents a pair of line.

Question 14.
Prove that the equation 2x² + 3xy – 2y² – x + 3y – 1 = 0 represents a pair of perpendicular straight lines.
Solution:
a = 2
b = -2
c = -1
f = \(\frac{3}{2}\)
g = –\(\frac{1}{2}\)
h = \(\frac{3}{2}\)
abc + 2fgh – af² – bg² – ch²
= 2(-2) (-1) + 2.\(\frac{3}{2}\) (-\(\frac{1}{2}\)).\(\frac{3}{2}\) – 2(\(\frac{9}{4}\)) + 2.\(\frac{1}{4}\) + \(\frac{1.9}{2}\)
= 4.-\(\frac{9}{4}\) – 2.\(\frac{9}{4}\) + \(\frac{1}{2}\) + \(\frac{9}{4}\)
= \(\frac{9}{2}\) – \(\frac{9}{2}\) = 0
h² – ab = \(\frac{9}{4}\) + 4 = \(\frac{25}{4}\) > 0,
g² – ac = \(\frac{1}{4}\) + 2 > 0
f² – bc = \(\frac{9}{4}\) – 2 = \(\frac{1}{4}\) > 0
a + b = 2 – 2 = 0
The given equation represents a pair of perpendicular line.

Question 15.
Show that the equation 2x² – 13xy – 7y² + x + 23y – 6 = 0 represents a pair of straight lines. Also find the angle between them and the co-ordinates of the point of intersection of the lines.
Solution:
Here a = 2
b = -7
c = -6
f = \(\frac{23}{2}\)
g = \(\frac{1}{2}\)
h = –\(\frac{13}{2}\)
abc + 2fgh – af² – bg² – ch²
= 2(-7) (-6) + 2.\(\frac{23}{2}\) . \(\frac{23}{2}\) . (-\(\frac{13}{2}\)) – 2\(\frac{529}{4}\) + 7.\(\frac{1}{4}\) + 6 . \(\frac{169}{4}\)
= \(\frac{1}{4}\) (336 – 299 – 1058 + 7 + 1014)
= \(\frac{1}{4}\) (1357 – 1357) = 0
h² – ab = \(\frac{169}{4}\) + 14 > 0,
g² – ac = \(\frac{1}{4}\) + 12 > 0,
f² – bc = \(\frac{529}{4}\) – 42 > 0
The given equation represents a pair of lines

= \(\frac{-13-92}{-56-169}\) = \(\frac{-105}{-225}\) = \(\frac{7}{15}\)
Point of intersection is P (\(\frac{19}{15}\), \(\frac{7}{15}\))

Question 16.
Find that value of λ for which the equation λx² – 10xy + 12y² + 5x – 16y -3 = 0 represents a pair of straight lines.
Solution:
Here a = λ
b = 12
c = -3
f = -8
g = \(\frac{5}{2}\)
h = -5
The given equation represents a pair of lines
abc + 2fgh – af² – bg² – ch² = 0
-36λ + 2(-8)\(\frac{5}{2}\) (-5) – λ . 64 – 12 . \(\frac{25}{4}\) + 3 . 25 = 0
-36λ + 200 – 64λ – 75 + 75 = 0
100λ = 200
⇒ λ = 2 ⇒ a = 2
h² – ab = 25 – 24 = 1 > 0
f² – bc = 64 + 36 = 100 > 0
g² – ac = \(\frac{25}{4}\) + 6 = \(\frac{49}{4}\) > 0
∴ The given equation represents a pair of lines for λ = 2.

Question 17.
Show that die pairs of straight lines 6x² – 5xy – 6y² = 0 and 6x² – 5xy – 6y² + x + 5y – 1 = 0 form a square.
Solution:
H ≅ 6x² – 5xy – 6y² = (3x + 2y) (2x – 3y)
and S ≅ 6x² – 5xy – 6y² + x + 5y – 1
= (3x + 2y- 1) (2x – 3y + 1).
∴ H = 0 represents the lines 3x + 2y = 0 and 2x – 3y = 0 which are perpendicular and S = 0 represents the lines 3x + 2y – 1 = 0, 2x – 3y + 1 = 0 which are also perpendicular. These two pairs of perpendicular lines, therefore, determine a rectangle.
Also the distance between the pair of opposite sides 3x + 2y = 0 and 3x + 2y – 1 = 0 is \(\frac{1}{\sqrt{13}}\) and this is the same as the distance between the other pair of opposite sides 2x – 3y = 0 and 2x – 3y + 1 = 0 of the rectangle. Hence the rectangle is a square.

Question 18.
Show that the equation 8x² – 24xy + 18y² – 6x + 9y – 5 = 0 represents a pair of parallel straight lines and find the distance between them.
Solution:
S = 8x² – 24xy + 18y² – 6x + 9y – 5
= 2(2x – 3y)² – 3(2x – 3y) – 5
= [2(2x – 3y) – 5] [(2x – 3y) + 1]
= (4x – 6y – 5) (2x – 3y + 1) = 0
The lines are 4x – 6y – 5 = 0 and 2x – 3y + 1 = 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{4}{2}\), \(\frac{b_{1}}{b_{2}}\) = \(\frac{-6}{-3}\) = 2
\(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\)
∴ The given equation represents a pair of parallel lines.
Distance between the lines
= 2\(\sqrt{\frac{g^{2}-a c}{a(a+b)}}\) = 2\(\sqrt{\frac{9+40}{8(8+18)}}\) = \(\frac{2.7}{2 \sqrt{52}}\) = \(\frac{7}{\sqrt{52}}\)

Question 19.
If the pair of lines represented by ax² + 2hxy + by² = 0 and ax² + 2hxy + by² + 2gx + 2fy + c = 0 form a rhombus, prove that (a – b) fg + h(f² – g²) = 0.
Solution:
Combined equation of OA and OB is ax² + 2hxy + by² = 0
Combined equation of AC, BC is ax² + 2hxy + by² + 2gx + 2fy + c = 0
Point of intersection is C(\(\frac{h f-b g}{a b-h^{2}}\), \(\frac{g h-a f}{a b-h^{2}}\))
Equation of the diagonal is y = \(\frac{g h-a f}{h f-b g}\) . x
y(hf – bg) = x(gh – af)
(gh – af) x – (hf- bg) y = 0

A and B are points on both this pair of lines.
Combined equation of AB is
2gx + 2fy + c = 0
OACB is a rhombus
OC and AB are perpendicular
2g(gh – af) – 2f(.hf – bg) = 0
hg² – afg – hf² + bfg = 0
(a – b) fg + h(f² – g²) = 0

Question 20.
If two of the sides of a parallelogram are represented by ax² + 2hxy + by² = 0 and px + qy = 1 is one of its diagonals, prove that the other diagonal is y(bp – hq) = x(aq – hp).
Solution:
Let OACB be the parallelogram two of whose sides \(\overleftrightarrow{O A}\), \(\overleftrightarrow{O B}\) are represented by the equation H ≡ ax² + 2hxy + by² = 0. Since the other pair of sides \(\overleftrightarrow{A C}\) and \(\overleftrightarrow{B C}\) are respectively parallel to \(\overleftrightarrow{O B}\) and \(\overleftrightarrow{O A}\), their combined equation will be of the form S ≡ ax² + 2hxy + by² + 2gx + 2fy + c = 0. Then the equation of the diagonal \(\overleftrightarrow{A B}\) is 2gx + 2fy + c = 0. But this line is given to be px + qy = 1 (or) -pcx – qcy + c = 0 where c ≠ 0

Therefore 2g = – pc, 2f = – qc …………….. (1)
The vertex C of the parallelogram = (\(\frac{h f-b g}{a b-h^{2}}\), \(\frac{g f-a f}{a b-h^{2}}\))
∴ Equation of the diagonal \(\overleftrightarrow{O C}\) is
(gh – af) x = (hf- bg)y
i.e., c(-ph + aq) x = c(-hq + bp)y (By 1)
(or) (aq – hp) x = (bp – hq) y (since c ≠ 0)

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B The Straight Line Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B The Straight Line Important Questions

Question 1.
Find the equation of the straight line passing through the point (2, 3) and making non-zero intercepts on the axes of coordinates whose sum is zero. [Mar 12]
Solution:
Equation of the line in the intercept form is
\(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
Given b = -a
Equation of the line is \(\frac{x}{a}\) – \(\frac{y}{a}\) = 1
⇒ x – y = a
This line passes through (2, 3)
2 – 3 = a ⇒ a = -1
Equation of the line is
x – y = -1 or x – y + 1 = 0

Question 2.
Find the equation of the straight line passing through the points (at₁², 2at₁ ) and (at₂², 2at₁ ).
Solution:
Equation of the given points.
(x – x₁) (y – y₂) = (y – y₁) (x₁ – x₂)
( x -at₁²) (2at₁ – 2at₂)
= (y – 2at₁) (at, — at2)
(x – at₁²) .2a(t₁ – t₂) = (y – 2at₁) a. (t₁² – t₂²)
2x – 2at₁² = y(t₁ + t₂) – 2at₁² + 2at₁t₂ = 0
2x – (t₁ + t₂)y + 2at₁t₂ = 0

Question 3.
Find the equation of the straight line passing through A (-1, 3) and i) parallel ii) perpendicular to the straight line passing through B(2, – 5) and C(4, 6). [Mar 11]
Solution:
Slope of BC = \(\frac{-5-6}{2-4}\) = \(\frac{-11}{-2}\) = \(\frac{11}{2}\)
i) The required line is parallel to BC and passes through A(- 1, 3)
Equation of the parallel line is
y – 3 = \(\frac{11}{2}\)(x + 1)
2y – 6 = 11x + 11
11x – 2y + 17 = 0

ii) The required line is perpendicular to BC.
Slope of the required line = –\(\frac{1}{m}\) = –\(\frac{2}{11}\)
This line passes through A (-1, 3)
Equation of the required line is
y – 3 = –\(\frac{2}{11}\)(x + 1)
11y – 33 = -2x – 2
2x + 11y – 31 = 0

Question 4.
Prove that the points (1, 11), (2, 15) and (-3, -5) are collinear and find the equation of the straight line containing them.
Solution:
A(1, 11), B(2, 15) and C (-3, -5) are the given points.
Equation of AB is
(y – y₁) (x₁ – x₂) = (x – x₁) (y₁ – y₂)
(y – 11) (1 -2) = (x – 1) (11 – 15)
-(y – 11) = -4 (x – 1)
-y + 11 = – 4x + 4
4x – y + 7 = 0
C (-3, -5) .
4x – y + 7 = 4(-3) + 5 + 7
= -12 + 12 = 0
C lines on AB ⇒ A, B, C are collinear.
Equation of the line containing them is 4x – y + 7 = 0

Question 5.
A straight line passing through A(1, -2) makes an angle tan^-1 \(\frac{4}{3}\) with the positive direction of the X – axis in the anticlock-wise sense. Find the points on the straight line whose distance from A is 5.
Solution:

Given α = tan^-1 \(\frac{4}{3}\) ⇒ tan α = \(\frac{4}{3}\)

cos α = \(\frac{3}{5}\), sin α = \(\frac{4}{5}\)
(x₁, y₁) = (1, -2) = x₁ = 1, y₁ = -2
Case i) .
r = 5
x = x₁ + r cos α = 1 + 5 . \(\frac{4}{5}\) = 1 + 4 = 5
y = y₁ + r sin α= – 2 + 5 . \(\frac{3}{5}\) = -2 + 3 = 1
Co-ordinates of B are (5, 1)

Case ii) r = -5
x = x₁ + r cos α = 1 – 5 . \(\frac{4}{5}\) = 1 – 4 = -3
y = y₁ + r sin α = -2 – 5 . \(\frac{3}{5}\) = -2 – 3 = -5
Co-ordinates of C are (- 3, – 5)

Question 6.
A straight line parallel to the line y = \(\sqrt{3} x\) passes through Q(2, 3) and cuts the line 2x + 4y – 27 = 0 at P. Find the length of PQ.
Solution:
PQ is parallel to the straight line y = \(\sqrt{3} x\)
tan α = \(\sqrt{3}\) = tan 60°
α = 60°
Q(2, 3) is a given point

Co-ordinates of any point P are
(x₁ + r cos α, y₁ + r sin α)
(2 + r cos 60°, 3 + r sin 60°)

Question 7.
Transform the equation
3x + 4y+ 12 = 0 into
i) slope—intercept form
ii) Intercept form and
in) normal form
Solution:
The given equation is 3x + 4y + 12=0
i) Slope-intercept form
4y = -3x – 12
y = (-\(\frac{3}{4}\))x + (-3)
Slope = –\(\frac{3}{4}\) , y – intercept = -3.

ii) Intercept form
-3x – 4y = 12
–\(\frac{3 x}{12}\) – \(\frac{4 y}{12}\) = 1
\(\frac{x}{(-4)}\) + \(\frac{y}{(-3)}\) = 1
x – intercept = -4, y – intercept = -3

iii) Normal form
-3x – 4y = 12
Dividing with \(\sqrt{9+16}\) = 5
(-\(\frac{3}{5}\))x + (-\(\frac{4}{5}\))y = \(\frac{12}{5}\)
Let cos α = \(\frac{-3}{5}\) and sin α = –\(\frac{4}{5}\)
p = \(\frac{12}{5}\) so that
x cos α + y sin α = p
α lies in third quadrant so that
α = π + tan^-1(\(\frac{4}{3}\))

Question 8.
If the area of the triangle formed by the straight line x = 0, y = 0 and 3x + 4y = a (a > 0) is 6, find the value of a. [May 11]
Solution:
Equation of the line is 3x + 4y = a

Given \(\frac{a^{2}}{2 a}\) = 6 ⇒ a² = 144
a = ±12
But a > 0
a = 12

Question 9.
Find the value of k, if the lines 2x – 3y + k = 0, 3x – 4y -13 = 0 and 8x – 11y – 33 = 0 are concurrent.
Solution:
Let L₁, L₂, L₃ be the straight lines whose equations are respectively
2x – 3y + k = 0 ………………. (1)
3x – 4y – 13 = 0 ……………….. (2)
8x – 11y – 33 = 0 ………………. (3)
Solving (2) and (3) for x and y

Point of the lines (2) and (3) is (11, 5)
The given lines L₁, L₂, L₃ are concurrent
∴ L₁ contain (11, 5)
∴ 2(11) – 3(5) + k = 0
k = -7

Question 10.
If the straight lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 are concurrent, then prove that a³ + b³ + c³ = 3abc.
Solution:
The equation of the given lines are
ax + by +c = 0 ………………… (1)
bx + cy + a = 0 ……………….. (2)
cx + ay + b = 0 ……………….. (3)
Solving (1) and (2) point of intersection is got by

c(ab – c²) +a (bc – a²) + b(ca – b²) = 0
abc – c³ + abc – a³ + abc – b³ = 0
∴ a³ + b³ + c³ = 3abc.

Question 11.
A variable straight line drawn through the point of intersection of the straight lines \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 and \(\frac{x}{b}\) + \(\frac{y}{a}\) = 1 meets the co-ordinate axes at A and B. Show that the locus of the mid point of \(\overline{\mathrm{AB}}\) is 2(a + b) xy = ab(x + y)
Solution:
Equations of the given lines are \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
and \(\frac{x}{b}\) + \(\frac{y}{a}\) = 1
Solving the point of intersection P(\(\frac{a b}{a+b}\), \(\frac{a b}{a+b}\))
Q (x₀, y₀) is any point on the locus
⇔ The line with x – intercept 2x₀, y – intercept 2y₀, passes through P
⇔ P lies on the straight line \(\frac{x}{2 x_{0}}\) + \(\frac{y}{2 y_{0}}\) = 1
i.e., \(\frac{a b}{a+b}\)(\(\frac{1}{2 x_{0}}\) + \(\frac{1}{2 y_{0}}\)) = 1
⇒ \(\frac{a b}{a+b}\) . \(\frac{x_{0}+y_{0}}{2 x_{0} y_{0}}\) = 0
ab(x₀ + y₀) = 2(a + b) x₀y₀
Q(x₀, y₀) lies on the curve
2(a + b)xy = ab(x + y)
Locks the mid of point of
AB is 2(a + b)xy = ab(x +y)

Question 12.
If a, b, c are in arithmetic progression, then show that the equation ax + by + c = 0 represents a family of concurrent lines and find the point of concurrency.
Solution:
a, b, c are in A.P.
2b = a + c
a – 2b + c = 0
a.1 + b(-2) + c = 0
Each number of family of straight lines
ax + by + c = 0
passes through the fixed point (1,-2)
∴ Set of lines ax + by + c = 0 for parametric values of a, b and c is a family of concurrent lines.
∴ Point of concurrency is (1, -2).

Question 13.
Find the value of k, if the angle between the straight lines 4x – y + 7 = 0 and kx – 5y – 9 = 0 is 45°.
Solution:

Squaring and cross multiplying
2(4k + 5)² = 17(k² + 25)
2(16k² + 40k + 25) = 17k² + 425
32k² + 80k + 50 = 17k² + 425
15k² + 80k – 375 = 0
3k² + 16k – 75 = 0
(k – 3) (3k + 25) = 0
k = 3 or -25/3

Question 14.
Find the equation of the straight line passing through (x₀, y₀) and (i) parallel (ii) perpendicular to the straight line
ax + by + c = 0.
Solution:
Equation of the given line is ax + by + c = 0
i) Equation of the parallel line is ax + by = k ………………… (1)
This line passes through P(x₀, y₀) ⇒ ax₀ + by₀ = k ……………… (2)
Subtracting (2) from (1) equation of the required line is a(x – x₀) + b(y – y₀) = 0

ii) Equation of the perpendicular line is bx – ay = k.
This line passes through P(x₀, y₀) ⇒ bx₀ – ay₀ = k
Subtracting, equation of the required line is b(x – x₀) – a(y – y₀) = 0

Question 15.
Find the equation of the straight line perpendicular to the line 5x – 2y = 7 and passing through the point of intersection of the lines 2x + 3y = 1 and 3x + 4y = 6.
Solution:
Given lines are L₁ = 2x + 3y – 1 = 0
L₂ = 3x + 4y – 6 = 0
Equation of the line passing through the intersection of L₁ = 0, L₂ = 0 is
L₁ + kL₂ = 0
(2x + 3y – 1) + k(3x + 4y – 6) = 0
(2 + 3k)x + (3 + 4k)y – (1 + 6k) = 0 …………………. (1)
This line is perpendicular to 5x – 2y = 7 ………………… (2)
a₁a₂ + b₁b₂ = 0
5(2 + 3k) – 2(3 + 4k) = 0
10 + 15k – 6 – 8k = 0
7k = -4 ⇒ k = – 4/7
Substituting in (1) equation of the required lines
(2 – \(\frac{12}{7}\))x + (3 – \(\frac{16}{7}\))x – (1 – \(\frac{24}{7}\)) = 0
\(\frac{2}{7}\)x + \(\frac{5}{7}\)y + \(\frac{17}{7}\) = 0 ⇒ 2x + 5y + 17 = 0

Question 16.
If 2x – 3y – 5 = 0 is the perpendicular bi-sector of the line segment joining (3, – 4) and (α, β) find α + β.
Solution:
(α, β) is the reflection of (3, -4) is the line
2x – 3y – 5 = 0
\(\frac{\alpha-3}{2}\) = \(\frac{\beta+4}{-3}\) = \(\frac{-2(6+12-5)}{4+9}\) = -2
α – 3 = -4 ⇒ α = -1
β + 4 = 6 ⇒ β = 2
α + β = -1 + 2 = 1

Question 17.
If the four straight lines ax + by + p = 0, ax + by + q = 0, cx + dy + r = 0 and cx + dy + s = 0 form a parallelogram, show that the area of the parallelogram so formed is \(\left|\frac{(p-a)(r-s)}{b c-a d}\right|\).
Solution:
Let L₁, L₂, L₃, L₄ be the lines given by
L₁ = ax + by + p = 0
L₂ = ax + by + q = 0
L₃ = cx + dy + r = 0
L₄ = cx + dy + s = 0
L₁ and L₂ are parallel: L₃ and L₄ are parallel
Area of the parallelogram = \(\frac{\mathrm{d}_{1} \mathrm{~d}_{2}}{\sin \theta}\)
d₁ = distance between L₁ and L₂ = \(\frac{|p-a|}{\sqrt{a^{2}+b^{2}}}\)
d₂ = distance between L₃ and L₄ = \(\frac{|r-s|}{\sqrt{c^{2}+d^{2}}}\)

Question 18.
The hypotenuse of a right angled isosceles triangle has its ends at the points (1, 3) and (-4, 1). Find the equations of the legs of the triangle.
Sol:
LetA =(1, 3)and B = (-4, 1) and ABC b a right isosceles triangle with \(\stackrel{\leftrightarrow}{A B}\) as hypotenuse.
We require, therefore, the equations of
\(\stackrel{\leftrightarrow}{A C}\) and \(\stackrel{\leftrightarrow}{B C}\)
Slope of \(\stackrel{\leftrightarrow}{A B}\) is \(\frac{1-3}{-4-1}\) = \(\frac{2}{5}\)
Since the slope of \(\stackrel{\leftrightarrow}{A B}\) is \(\), neither \(\frac{2}{5}\) nor \(\stackrel{\leftrightarrow}{A C}\) is vertical.

Taking the slope of \(\stackrel{\leftrightarrow}{A C}\) as \(\frac{7}{3}\), the slope of \(\stackrel{\leftrightarrow}{B C}\) would be –\(\frac{3}{7}\). Therefore, the equations of \(\stackrel{\leftrightarrow}{A C}\) and \(\stackrel{\leftrightarrow}{B C}\) are respectively.
y – 3 = \(\frac{7}{3}\) (x – 1) and y – 1 = – \(\frac{3}{7}\) (x + 4),
which become 7x – 3y + 2 = 0 and 3x + 7y + 5 = 0
If the lines drawn through A and B respectively parallel to \(\stackrel{\leftrightarrow}{B C}\) and \(\stackrel{\leftrightarrow}{A C}\) meet at D, then ∆ABD is also right isosceles, having \(\stackrel{\leftrightarrow}{A B}\) as its hypotenuse.
Therefore, the equations of \(\stackrel{\leftrightarrow}{A D}\) and \(\stackrel{\leftrightarrow}{B D}\) are respectively,
3(x – 1) + 7(y – 3) = 0 and 7(x + 4) – 3(y – 1) = 0
⇒ 3x + 7y – 24 = 0 and 7x – 3y + 31 = 0.
Therefore, the two pairs of legs required are
7x – 3y + 2 = 0, 3x + 7y + 5 = 0 and
3x + 7y – 24 = 0, 7x – 3y + 31 = 0.
Note : ADBC is square.

Question 19.
A line is such that its segment between the lines 5x – y + 4 = 0 and 3x + 4y – 4 = 0 is . bisected at the point (1, 5). Obtain its equation.
Solution:
Let the required line meet 3x + 4y – 4 = 0 at A and 5x – y + 4 = 0 at B, so that AB is the segment between the given lines, with its mid-point at C = (1, 5).
The equation 5x – y + 4 = 0 can be written as
y = 5x + 4 so that any point on \(\stackrel{\leftrightarrow}{B X}\) is (t, 5t + 4) for all real t.
∴ B = (t, 5t + 4) for some t. Since (1, 5) is the mid-point of \(\stackrel{\leftrightarrow}{A B}\)
A = [2 – t, 10 – (5t + 4)]
= [2 – t, 6 – 5t]
Since A lies on 3x + 4y – 4 = 0,
3(2 – 1) + 4(6 – 5t) – 4 = 0
⇒ -23t + 26 = 0
⇒ t = \(\frac{26}{23}\)

equation of \(\stackrel{\leftrightarrow}{A B}\) is y – 5 = \(\frac{107}{3}\) (x – 1)
⇒ 3y – 15 = 107x – 107
⇒ 107x – 3y – 92 = 0.

Question 20.
An equilateral triangle has its incentre at the origin and one side as x + y – 2 = 0. Find the vertex opposite to x + y – 2 = 0.
Solution:
Let ABC be the equilateral triangle and
x + y – 2 = 0 represent side \(\stackrel{\leftrightarrow}{B C}\).
Since O is the incentre of the triangle, \(\stackrel{\leftrightarrow}{A D}\) is the bisector of ∠BAC .
Since the triangle is equilateral, \(\stackrel{\leftrightarrow}{A D}\) is the perpendicular bisector of \(\stackrel{\leftrightarrow}{B C}\).
Since O is also the centroid, AO : OD = 2 : 1. [The centroid, circumcentre incentre and orthocentre coincide]
Let D = (h, k).
Since D is the foot of the perpendicular from O onto \(\stackrel{\leftrightarrow}{B C}\), D is given by

∴ x₁ = -2, y₁ = -2.
∴ A = (-2, -2), the required vertex.

Question 21.
Find the orthocentre of the triangle whose vertices are (-5,-7) (13, 2) and (-5, 6)
Solution:
Let A(-5, -7), B(13,2) and C (-5, 6) be the vertices of a triangle. Let \(\stackrel{\leftrightarrow}{A D}\) be the perpendicular drawn from A to \(\stackrel{\leftrightarrow}{B C}\) and \(\stackrel{\leftrightarrow}{B E}\) be the perpendicular drawn from B to \(\stackrel{\leftrightarrow}{A C}\).

Now slope of \(\stackrel{\leftrightarrow}{B C}\) = \(\frac{6-2}{-5-13}\) = \(\frac{-2}{9}\)
Since \(\stackrel{\leftrightarrow}{A D}\) ⊥ \(\stackrel{\leftrightarrow}{B C}\) , slope of \(\stackrel{\leftrightarrow}{A D}\) = \(\frac{9}{2}\) and so, the equation of \(\stackrel{\leftrightarrow}{A D}\) is
9x – 2y = – 45 + 14 = -31 ………………… (1)
Equation of \(\stackrel{\leftrightarrow}{A C}\) is x = -5 which is a vertical line and therefore equation of \(\stackrel{\leftrightarrow}{B E}\) is y = 2. ……………… (2)
Point of intersection of the lines (1) and (2) is (-3, 2) which is the orthocentre of ∆ ABC.

Question 22.
If the equations of the sides of a triangle are 7x + y – 10 = 0, x – 2y + 5 = 0 and x + y + 2 = 0, find the orthocentre of the triangle.
Solution:
Let ∆ ABC be the given triangle
Let the equations x – 2y + 5 = 0 ……………… (1)
7x + y – 10 = 0 ……………… (2)
and x + y + 2 = 0 ……………….. (3)
represent the sides \(\overleftrightarrow{\mathrm{AB}}\), \(\overleftrightarrow{\mathrm{BC}}\), and \(\overleftrightarrow{\mathrm{CA}}\)

Let \(\overleftrightarrow{\mathrm{AD}}\) and \(\overleftrightarrow{\mathrm{BE}}\) be the altitudes drawn from A and B respectively to the sides \(\overleftrightarrow{\mathrm{BC}}\) & \(\overleftrightarrow{\mathrm{CA}}\)
Solving the equations (1) and (3), we obtain A =(-3,1).
Since \(\overleftrightarrow{\mathrm{AD}}\) ⊥ \(\overleftrightarrow{\mathrm{BC}}\), the equations of \(\overleftrightarrow{\mathrm{AD}}\) is
x – 7y = -3 – 7 = -10 …………………… (4)
Solving the equation (1) and (2), we obtain B = (1, 3).
Since \(\overleftrightarrow{\mathrm{BE}}\) ⊥ \(\overleftrightarrow{\mathrm{AC}}\) , the equation of BE is
x – y = 1 – 3 = -2 …………… (5)
Point of intersection of the lines (4) and (5) is H(\(\frac{-2}{3}\), \(\frac{4}{3}\)) which is the orthocentre of ∆ ABC.

Question 23.
Find the circumcentre of the triangle whose vertices are (1, 3), (-3, 5) and (5, -1).
Solution:
Let the vertices of the triangle be
A(1, 3), B(-3, 5) and (5,-1).
The midpoints of the sides \(\overline{\mathrm{BC}}\), \(\overline{\mathrm{CA}}\) are respectively D(1, 2) and E(3, 1).
Let S be the point of intersection of the perpendicular bisectors of the sides \overline{\mathrm{BC}} and \overline{\mathrm{CA}}

Since the slope of \(\overleftrightarrow{\mathrm{BC}}\) = \(\frac{5+1}{-3-5}\) = \(\frac{-3}{4}\), the slope \(\overleftrightarrow{\mathrm{SD}}\) is \(\frac{4}{3}\) and therefore its equation is 4x – 3y = 4 – 6 = -2 …………….. (1)
Slope of \(\overleftrightarrow{\mathrm{AC}}\) = \(\frac{3+1}{1-5}\) = -1 ⇒ Slope of –\(\sqrt{3x}\) = 1
∴ Equation of \(\overleftrightarrow{\mathrm{SE}}\) is x – y = 3 – 1 = 2 ………………….. (2),
Solving the equations (1) and (2),we obtain S = (-8, -10) which is the circumcentre of ∆ ABC.

Question 24.
Find the circumcentre of the triangle whose sides are 3x – y – 5 = 0, x + 2y – 4 = 0 and 5x + 3y + 1 = 0. [May 11, 05; Mar. 06]
Solution:
Let the given equations 3x – y – 5 = 0, x + 2y – 4 = 0 and 5x + 3y + 1 = 0 represent the sides \(\overleftrightarrow{\mathrm{BC}}\), \(\overleftrightarrow{\mathrm{CA}}\) and \(\overleftrightarrow{\mathrm{AB}}\) respectively of ∆ ABC.
Solving the above equations two by two, we obtain the vertices A(-2, 3), B(1, -2) and (2, 1) of the given triangle.
The midpoints of the sides \(\overline{\mathrm{BC}}\) and \(\overline{\mathrm{CA}}\) are respectively D = (\(\frac{3}{2}\), \(\frac{-1}{2}\)) and E = (0, 2).

Equation of \(\overleftrightarrow{\mathrm{SD}}\) the perpendicular bisector of \(\overline{\mathrm{BC}}\) is x + 3y = 0 and the equation of \(\overleftrightarrow{\mathrm{SE}}\) the perpendicular bisector of \(\overline{\mathrm{AC}}\) is 2x – y + 2 = 0. Solving these two equations, we obtain the point of intersection of the lines \(\overleftrightarrow{\mathrm{SD}}\) and \(\overleftrightarrow{\mathrm{SE}}\) which is S(\(\frac{-6}{7}\), \(\frac{2}{7}\)) the circumcentre of ∆ ABC.

Question 25.
Find the incentre of the triangle formed by the straight lines y = \(\sqrt{3}\)x, y = –\(\sqrt{3}\)x and y = 3.
Solution:

The straight lines y = \(\sqrt{3x}\) and y = –\(\sqrt{3x}\) respectively make angles 60° and 120° with the positive direction of X – axis.
Since y = 3 is a horizontal line, the triangle formed by the three given lines is equilateral. So in-centre is same and centroid.
Vertices of the triangle and O(0, 0), A(\(\sqrt{3}\), 3) and D(\(\sqrt{3}\), 3).
∴ Incentre is (\(\frac{0+\sqrt{3}-\sqrt{3}}{3}\), \(\frac{0+3+3}{3}\)) = (0, 2)

Question 26.
Find the equation of the straight line whose distance from the origin is 4, If the normal ray from the origin to the straight line makes an angle of 135° with the positive direction of the x-axis.
Solution:
The equation of the given line is
x cos α + y sin α = p where p = 4 and α = 135°
∴ x(\(\frac{-1}{\sqrt{2}}\)) + y(\(\frac{1}{\sqrt{2}}\)) = 4 ⇒ or x – y + 4\(\sqrt{2}\) = 0

(i). Transform the equation x + y + 1 = 0 in to normal from.
Solution:
x + y + 1 = 0
⇔ (\(\frac{-1}{\sqrt{2}}\))n + (\(\frac{-1}{\sqrt{2}}\)) y = \(\frac{1}{\sqrt{2}}\)
⇔ x cos \(\frac{5\pi}{42}\) + y sin \(\frac{5\pi}{4}\) = \(\frac{1}{\sqrt{2}}\)
Hence the normal form of the equation of the given straight line is x cos \(\frac{5\pi}{4}\) + y sin \(\frac{5\pi}{4}\) = \(\frac{1}{\sqrt{2}}\) and the distance of the line from the origin is = \(\frac{1}{\sqrt{2}}\)

ii) A straight line passing through A(1, -2) makes and angle Tan^-1\(\frac{4}{3}\) with the positive direction of the x – axis in the anti-clock wise sense. Find the points of the straight line whose distance from A is 5.
Solution:
The paramatic equation of the lines A (1, -2)
Slope is \(\frac{4}{3}\) (∵ tan θ = \(\frac{4}{3}\))
are x = 1 + r cos θ = 1 + r(\(\frac{3}{5}\)) and y
= -2 + r sinθ
= -2 + r (\(\frac{4}{5}\))
The points on the above line a + a distance of (r) = 5 correspond to r = ±5 in the above equation and are
∴ (4, 2) and (-2, -6)

Question 27.
Trans form the equation 3x + 4y + 12 = 0 in to (i) slope – intercept from (ii) intercept form (iii) normal form.
Solution:
(i) slop – intercept form
3x + 4y + 12 = 0
⇔ 4y = -3x – 12
⇔ y = (\(\frac{-3}{4}\)) x + (-3)
∴ Slope = (\(\frac{-3}{4}\)) and y – intercept = -3,

ii) Intercept form :
3x = 4y + 12 = 0
⇔ \(\frac{-3y}{12}\) – \(\frac{-4y}{12}\) = 1
⇔ \(\frac{x}{(-4)}\) + \(\frac{y}{(-3)}\) = 1
∴ x-intercept of the line is -4, and tan y-intercept is -3.

iii) Normal form
3x = 4y + 12 = 0
⇔ -3x + -4y – 12 = 0
⇔ (\(\frac{-3}{5}\))x + (\(\frac{-4}{5}\))y = \(\frac{12}{5}\)
⇔ x cos α + y sin α = p ⇒ p\(\frac{12}{5}\) and cos α = \(\frac{-3}{5}\), sin α = \(\frac{-4}{5}\) determine the angle α in (0, 2π).

Question 28.
Find the angle between the lines 2x + y + 4 = 0 and y – 3x = 7.
Solution:
The angle between the given lines
= cos^-1 \(\frac{-6+1}{\sqrt{5 \times 10}}\)
= cos^-1 [latex]\frac{5}{\sqrt{2}}[/latex] = cos^-1(\(\frac{1}{\sqrt{2}}\)) = \(\frac{\pi}{4}\)

Question 29.
Find Q(h,k) in the foot of the perpendicular from p(x₁, y₁) on the straight lines ax + by + c = 0 then (h – x₁) ; a = (k – y₁); b = -(ax₁ + by₁ + c); (a² + b²)
Solution:
Equation of \(\stackrel{\leftrightarrow}{P Q}\) which is normal to the given straight line
L : ax + by + c = 0
bx – ay = bx₁ – ay₁

∴ Q ∈ \(\stackrel{\leftrightarrow}{P Q}\) we have
bh – ak = bx₁ – ay₁
∴ b(h – x₁) = a(k – y₁)
or (h – x₁) a = (k – y₁); b.
But, this implus that h = aλ + x₁, and k = bλ + y₁
for some λ ∈ R. sine Q(h, k) in point on L.
a(aλ + x₁) + b(bλ + y₁) + c = 0
i.e., λ = \(\frac{\left(a x_{1}+b y_{1}+c\right.}{\left(a^{2}+b^{2}\right)}\)
∴ (h – x₁) ; a = (k – y₁); b
= – (ax₁ + by₁ + c); (a² + b²)

Question 30.
Find the distance between the parallel straight lines 3x + 4y – 3 = 0 and 6x + 8y -1 = 0.
Solution:
The equation of the given straight lines is 3x + 4y – 3 = 0 and 6x + 8y – 1 = 0.
formula : \(\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\) = \(\frac{-6+1}{\sqrt{6^{2}+8^{2}}}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\)

Question 31.
Find the condition for the points (a, o) (h, k) and (o, b) when ab ≠ 0 to be collinear. [Mar 10]
Solution:
The equation of the line passing through (a, o) and (o, b) is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
The given points are collinear ⇒ (h, k) lies on the above line ⇒ \(\frac{h}{a}\) + \(\frac{k}{b}\) = 1
⇒ hb + ka = ab

Question 32.
Find the area of the triangle formed by the straight lines x cos a + y sin a = p and the axes of co-ordinates. [Mar 10]
Solution:
The area of the triangle formed by the line ax + by + c = 0
and the co-ordinate axes is \(\frac{c^{2}}{2|a b|}\)
∴ Area of the triangle = \(\frac{p^{2}}{2|\cos \alpha \cdot \sin \alpha|}\)
= \(\frac{p^{2}}{|\sin 2 \alpha|}\)

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Transformation of Axes Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B Transformation of Axes Important Questions

Question 1.
When the origin is shifted to (2, 3) by the translation of axes, the coordinates of a point p are changed as (4, -3). Find the coordinates of P in the original system.
Solution:
(h, k) = (2, 3) ⇒ h = 2, k = 3
(x’, y’) = (4, 3) ⇒ x = 4, y = -3
x = x’ + h = 4 + 2 = 6, y = y’ + k = -3 + 3 = 0
original Co-ordinates are (6, 0)

Question 2.
Find the point to which the origin is to be shifted by the translation of axes so as to remove the first degree terms from the equation
ax² + 2hxy + by² + 2gx + 2fy + c = 0, where h² ≠ ab.
Solution:
Let the origin be shifted to (α, β) by the translation of axes.
Then x = x’ + α,
y = y’ + β
On substituting these in the given equation, we get
a(x’ + α)² + 2h(x’ + α) (y’ + β) + b(y’ + β)² + 2g(x’ + α) + 2f(y’ + β) + c = 0
Which gives
ax’² + 2hx’y’ + by² + 2x'(α + β + g) + 2y'(hα + bβ + f) + aα² + 2hαβ + bβ² + 2gα + 2fβ + c = 0 ……….. (1)
If equation (1) has to be free from the first degree terms, then we have
aα + hβ + g = 0 and hα + bβ +1 = 0
Solving these equations for a and 13, we get
α =\(\frac{h f-b g}{a b-h^{2}}\) , β = \(\frac{g h-a f}{a b-h^{2}}\)
Therefore, the origin is to be shifted to
\(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)

Question 3.
Find the point to which the origin is to be shifted by the translation of axes so as to remove the first degree terms from the equation
ax² + by² + 2gx + 2fy + c = 0, where a ≠ 0, b ≠ 0.
Solution:
Here the given equation does not contain xy term. Hence writing h = 0 in the result of
Problem 12, the required point is \(\left(\frac{-g}{a}, \frac{-f}{b}\right)\)

Question 4.
If the point P changes to (4, -3) when the axes are rotated through an angle of 135°, find the coordinates of P with respect to the original system.
Solution:
Here (x’, y’) = (4, -3); θ = 135°
Let (x, y) be the coordinates of P, then
x = x’ cos θ – y’ sin θ
= 4cos 135° – (-3) sin 135°

Therefore, the coordinates of P with respect to the original system are (\(\frac{-1}{\sqrt{2}}\), \(\frac{7}{\sqrt{2}}\))

Question 5.
Show that the axes are to be rotated through an angle of \(\frac{1}{2}\) Tan^-1(\(\frac{2 h}{a-b}\)) so as to remove the xy term from the equation ax² + 2hxy + by² = 0, if a ≠ b and through the angle \(\frac{\pi}{4}\), if a = b. [Mar 13, 06]
Solution:
If the axes are rotated through an angle ‘θ’,
then
x = x’ cos θ – y’ sin θ
y = x’ sin θ + y’ cos θ
Therefore the given equation transforms as
a(x’ cos θ – y’ sin θ)² + 2h(x’ cosθ – y’ sinθ) (x’ sinθ + y’ cosθ) + b(x’ sin θ + y’ cos θ)² = o
To remove x’y’ term from the equation, the coefficient of x’y’ term must be zero.
So, (b – a) sin θ cos θ + h(cos²θ – sin²θ) = 0
i.e., h cos 2θ = \(\frac{a-b}{2}\) sin 2θ
i.e., tan 2θ = \(\frac{2 h}{a-b}\), if a ≠ b and h cos 2θ = 0, if a = b
Therefore θ = \(\frac{1}{2}\) Tan^-1(\(\frac{2 h}{a-b}\)), if a ≠ b and θ = \(\frac{\pi}{4}\), If a = b

Question 6.
When the origin is shifted to (-2, -3) and the axes are rotated through an angle 45° find the transformed equation of 2x² + 4xy – 5y² + 20x – 14 = 0.
Solution:
Here (h, k) = (-2, -3), h = -2, k = -3
θ = 45°
Let (x’, y’) be the new co-ordinates of any point (x, y) is the plane after transformation
x = x’ cos θ – y’ sin θ + h = -2 + x’ cos 45° – y’ sin 45°.
= -2 + \(\frac{x^{\prime}-y^{\prime}}{\sqrt{2}}\)
y = x’ si nθ + y’ cos θ + k = x’ sin 45° + y’ cos
45° – 3 = -3 + \(\frac{x^{\prime}+y^{\prime}}{\sqrt{2}}\)
The transformed equation is

⇒ (x’ – y’)² + 8 – 4\(\sqrt{2}\) (x’ – y’) + 2(x’² – y’²) – 6\(\sqrt{2}\) (x’ – y’) – 4\(\sqrt{2}\) (x’ + y’) + 24 = 0
⇒ –\(\frac{5}{2}\) (x’ – y’)² – 45 + 15\(\sqrt{2}\) (x’ + y’) + 10\(\sqrt{2}\) (x’ – y’) – 40 – 11\(\sqrt{2}\) (x’ + y’) + 66 – 14 = 0
⇒ x’² + y’² – 2x’y’ + 2x’² – 2y’² – \(\frac{5}{2}\) (x’² + y’² + 2x’y’) – 1 = 0
\(\frac{1}{2}\) x’² – \(\frac{7}{2}\) y’² – 7x’y’ – 1 = 0
i.e., x’² – 7y’² – 14x’y’ – 2 = 0
The transformed equation is (dropping dashes)
x² – 7y² – 14xy – 2 = 0

Question 7.
When the origin is shifted to (-2, 3) by translation of axes, let us find the co-ordinates of (1, 2) with respect to new axes.
Solution:
Here (h, k) = (-2, 3)
Let (x, y) = (1, 2) be shifted to (x’ y’) by the translation of axes.
Then (x’, y’) = (x – h, y – k)
= (1 – (-2), 2 – 3) = (3, -1)

Question 8.
When the origin is shifted to (3, 4) by the translation of axes, let us find the transformed equation of 2x² + 4xy + 5y² = 0.
Solution:
Here (h, k) = (3, 4)
on substituting x = x’ + 3 and y = y’ + 4 in the given equation.
as per note the equation f(x, y) = 0 of the curve is transformed as f(x’ + h, y’ + k) = 0]
we get
2(x’ + 3) + 4(x’ + 3) (y’ + 4) + 5(y’ + 4)² = o
Simplifying this equation, we get
2x’² + 4x’y’ + 5y’² + 28x’ + 52y’ + 146 = 0
This equation can be written (dropping dashes) as
2x² + 4xy + 5y² + 28x + 52y + 146 = 0

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Locus Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B Locus Important Questions

Question 1.
Find the equation of the locus of a point which is at a distance 5 from (-2, 3) in the xoy plane.
Solution:
Let the given point be A = (-2, 3) and P(x, y) be a point on the plane.
The geometric condition to be satisfied by P to be on the locus is that
AP = 5 …………… (1)
Expressing this condition algebraically, we get
\(\sqrt{(x+2)^{2}+(y-3)^{2}}\) = 5
i.e., x² + 4x + 4 + y² – 6y + 9 = 25
i.e., x² + y² + 4x – 6y – 12 = 0 …………….. (2)
Let Q(x₁, y₁) satisfy (2).
Then, x₁² + y₁² + 4x₁ – 6y₁ – 12 = 0 …………. (3)
Now the distance of A from Q is
AQ = \(\sqrt{\left(x_{1}+2\right)^{2}+\left(y_{1}-3\right)^{2}}\)
∴ AQ² = x₁² + 4x₁ + 4 + y₁² – 6y₁ + 9
= (x₁² + y₁² + 4x₁ – 6y₁ – 12) + 25
= 25 (by using (3))
Hence AQ = 5.
This means that Q(x₁, y₁) satisfies the geometric condition (1).
∴ The required equation of locus is
x² + y² + 4x – 6y – 12 = 0.

Question 2.
Find the equation of locus of a point P, if the distance of P from A(3, 0) is twice the distance of P from B(-3, 0).
Solution:
Let P(x, y) he a point on the locus. Then the geometric condition to be satisfied by P is
PA = 2PB …………….. (1)
i.e., PA² = 4PB²
i.e., (x – 3)² + y² = 4[(x + 3)² + y²]
i.e., x² – 6x + 9 + y² = 4(x² + 6x + 9 + y²]
i.e., 3x² + 3y² + 30x + 27 = 0
i.e., x² + y² + 10x + 9 = 0 ………………. (2)
i.e., Q(x₁, y₁) satisfy (2).
Then x₁² + y₁² + 10x₁ + 9 = 0 …………….. (3)
Now QA² = (x₁ – 3)² + y₁²
= x₁² – 6x₁ + 9 + y₁²
= 4x₁² + 24x₁ + 36 + 4y₁² – 3x₁² – 30x₁ – 27 – 3y₁²
= 4(x₁² + 6x₁ + 9 + y₁²) – 3(x₁² + 10x₁ + 9 + y₁²)
= 4(x₁² + 6x₁ + 9 + y₁²) (by using(3))
= 4 [(x₁ + 3)² + y₁²]
= 4QB²
∴ QA = 2QB.
This means that Q(x₁, y₁) satisfies (1).
Hence, the required equation of locus is
x² + y² + 10x + 9 = 0.

Question 3.
Find the locus of the third vertex of a right angled triangle, the ends of whose hypotenuse are (4, 0) and (0, 4).
Solution:
Let A = (4, 0) and B = (0, 4).
Let P(x, y) be a point such that. PA and PB are perpendicular. Then PA² + PB² = AB².
i.e., (x – 4)² + y² + x² + (y – 4)² = 16 + 16
i.e., 2x² + 2y² – 8x – 8y = 0
or x² + y² – 4x – 4y = 0
Let Q(x₁, y₁) satisfy (2) and Q be different from A and B.
Then x₁² + y₁² — 4x₁ – 4y₁ = 0,
(x₁, y₁) ≠ (4, 0) and (x₁, y₁) ≠ (0, 4) ……………… (3)
Now QA² + QB² = (x₁ – 4)² + y₁² + x₁² + (y₁ – 4)²
= x₁² – 8x₁ + 16 + y₁² + x₁² + y₁² – 8y₁ + 16
= 2(x₁² + y₁² – 4x₁ – 4y₁) + 32
= 32 (by using (3))
= AB²
Hence QA² + QB² = AB², Q ≠ A and Q ≠ B.
This means that Q(x₁, y₁) satisfies (1).
∴ The required equation of locus is (2),which is the circle with \(\overline{\mathrm{AB}}\) as diameter, deleting the points A and B.
Though A and B satisfy equation (2), they do not satisfy the required geometric condition.

Question 4.
Find the equation of the locus of P, if the ratio of the distances from P to A(5, -4) and B(7, 6) is 2 : 3. [Mar 14]
Solution:
Let P(x, y) be any point on the locus.
The geometric condition to be satisfied by P
is \(\frac{\mathrm{AP}}{\mathrm{PB}}\) = \(\frac{1}{2}\)
i.e., 3AP = 2PB …………….. (1)
i.e., 9AP² = 4PB²
i.e., 9[(x – 5)² + (y + 4)²] = 4[(x – 7)² + (y – 6)²]
i.e., 9[x² + 25 – 10x + y² + 16 + 8y]
= 4[x² + 49 – 14x + y² + 36 – 12y]
i.e., 5x² + 59 – 34x + 120y + 29 = 0 ……………….. (2)
Let Q(x₁, y₁) satisfy (2). Then
5x₁² + 5y₁² – 34x₁ + 120y₁ + 29 = 0 ……………….. (3)
Now 9AQ² = 9[x₁² + 25 – 10x₁ + y₁² + 16 + 8y₁]
= 5x₁² + 5y₁² – 34x₁ + 120y₁ + 29 + 4x₁² + 4y₁² – 56x₁ – 48y₁ + 340
= 4[x₁² + y₁² – 14x₁ – 12y₁ + 49 + 36] (by using (3))
= 4[(x₁ – 7)² + (y₁ – 6)²] = 4PB²
Thus 3AQ = 2PB. This means that Q(x₁, y₁) satisfies (1).
Hence, the required equation of locus is
5(x² + y²) – 34x + 120y + 29 = 0.

Question 5.
A(2, 3) and B(-3, 4) are two given points. Find the equation of locus of P so that the area of the triangle PAB is 8.5. [Mar 11]
Solution:
Let P(x, y) be a point on the locus.
The geometric condition to be satisfied by P is that,
area of ∆PAB = 8.5 …………………(1)
i.e., \(\frac{1}{2}\)|x(3 – 4) + 2(4 – y) – 3(y – 3)| = 8.5
i.e., |-x + 8 – 2y – 3y + 9| = 17
i.e., |-x – 5y + 17| = 17
i.e., -x – 5y + 17 = 17 or -x – 5y + 17 = -17
i.e., x + 5y = 0 or x + 5y = 34
∴ (x + 5y) (x + 5y – 34) = 0
i.e., x² + 10xy + 25y² – 34x – 170y = 0 …………… (2)
Let Q(x₁, y₁) satisfy (2). Then
x₁ + 5y₁ = 0 or x₁ + 5y₁ = 34 ……………………. (3)
Now, area of ∆QAB
= \(\frac{1}{2}\)|x₁(3 – 4) + 2(4 – y₁) – 3(y₁ – 3)|
= \(\frac{1}{2}\)|-x₁ + 8 – 2y₁ – 3y₁ + 9|
= \(\frac{1}{2}\) |-x₁ – 5y₁ + 17|
= \(\frac{17}{2}\) = 8.5 (by using (3))
This means that Q(x₁, y₁) satisfies (1).
Hence, the required equation of locus is
(x + 5y)(x + 5y – 34) = 0 or
x² + 10xy + 25y² – 34x – 170y = 0.

Use these Inter 1st Year Maths 1B Formulas PDF Chapter 10 Applications of Derivatives to solve questions creatively.

Intermediate 1st Year Maths 1B Applications of Derivatives Formulas

→ Error in y = Δy = f(x + Δx) – f(x)

→ Differential in y = dy = f’ (x) Δx

→ Relative error in y = \(\frac{\Delta y}{y}\)

→ Percentage error in y = \(\frac{\Delta y}{y}\) × 100

Circle:
if’ r’ is the radius, x is the diameter, p is the perimeter (circumference) and A is the area of a circle then

• A = πr or A = \(\frac{\pi x^{2}}{4}\)
• x = 2r
• p = 2πr (or) p = πx

Sector :
If ‘r’ is the radius, l is the length of arc and ‘θ’ is the angle then

• Area = A = \(\frac{1}{2}\)r (or) A = \(\frac{1}{2}\)r²θ
• Perimeter = p = l + 2r (or) p = r (θ + 2)
• l = rθ

Cylinder :
If ‘r’ is the radius, h is the height then

• Lateral surface area = 2πrh
• Total surface area = S = 2πrh + 2πr²
• Volume = V = πr²h

Cone:

r is base radius, h is the height, ‘l’ is the slant height, a is the vertical angle then

• l² = r² + h²
• tan \(\frac{\alpha}{2}=\frac{r}{h}\)
• Lateral surface area = πrl (or) πr\(\sqrt{r^{2}+h^{2}}\)
• Total surface area = S = πrl + πr² (or) S = πr \(\sqrt{r^{2}+h^{2}}\) + πr²
• Volume = V = \(\frac{1}{3}\)πr²h

Simple pendulum:
If ‘l’ is the length, T is the period of oscillation of a simple pendulum and g is the acceleration due to gravity then T = 2π\(\sqrt{\frac{l}{g}}\)

Sphere :
‘r’ is the radius,

• Surface area = S = 4πr
• Volume = V = \(\frac{4}{3}\) πr³

Cube :
Let ‘x’ is the side

• Surface area = S = 6x²
• Volume = V = x³

→ Slope of the tangent = f'(x)

→ Equation of the tangent is (a, b) is y – b = f'(a) (x – a)

→ Equation of the normal at (a, b) is y – b = –\(\frac{1}{f^{\prime}(a)}\) (x – a)

→ Length of the tangent = \(\left|\frac{f(a) \sqrt{1+\left(f^{\prime}(a)\right)^{2}}}{f^{\prime}(a)}\right|\)

→ Length of the normal = \(\left|f(a) \sqrt{1+\left[f^{*}(a)\right]^{2}}\right|\)

→ Length of the sub-tangent = \(\left|\frac{f(a)}{f^{\prime}(a)}\right|\)

→ Length of the sub-normal = |f(a).f'(a)|

Angle between the curves is tan Φ = \(\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|\)

→ If m₁ = m₂, the curves touch each other and have a common tangent

→ If m₁m₂ = -1, the curves cut orthogonally

→ If a particle starts from a fixed point and moves a distance ‘s’ along a straight line during time ‘t’, then
v = velocity of the particle at the time t = \(\frac{\mathrm{ds}}{\mathrm{dt}}\)
a = acceleration of the particle at the time t = \(\frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}\) = v\(\frac{\mathrm{dv}}{\mathrm{ds}}\)

• If v > 0, then the particle is moving away from the straight point.
• If v < 0, then the particle is moving towards the straight point.
• If v = 0, then the particle comes to rest.

→ Let O be a fixed in a plane and OX be a fixed ray in the same plane. Let P be the position of a particle on a curve C, at time ‘f and ∠XOP = θ.

• ω = Angular velocity of the particle around ‘O’ = \(\frac{\mathrm{d} \theta}{\mathrm{dt}}\)
• Angular acceleration of the particle around ‘O’ = \(\frac{\mathrm{d}^{2} \theta}{\mathrm{dt}{ }^{2}}\) (Where θ is in radian measure).

→ Increasing and Decreasing functions: A function f(x) is

• Increasing if f'(x) > 0
• Decreasing if f'(x) < 0′
• f(x) is stationary if f’ (x) = 0

→ A function f(x) has local maxima or local minima only at stationary points.

• f(x) has local maximum of f’(x) = 0, f”(x) < 0 f(x) has local minimum if
• f’(x) = 0, f”(x) > 0
  Absolute maximum = max. {local maxima}
  Absolute minimum = Min {local minima}.

Tangent To A Curve:
Definition :
Let y = f(x) be a curve and P be a point on the curve. If Q is a point on the curve other than P, then PQ is called a secant line of the curve. If the secant line \(\overline{P Q}\) approaches the same limiting position as Q approaches P along the curve from either side then the limiting position is called the tangent line to the curve at the point P. The point P is called the point of contact of the tangent line to the curve.

The tangent at a point to a curve, if it exists, is unique. Therefore, there exists at most one tangent at a point to a curve.

Geometrical Interpretation Of Derivative:
Let P be a point on the curve y = f(x). Then the slope of the tangent to the curve at P is equal to \(\left(\frac{d y}{d x}\right)_{P}\) i.e., (f'(x))_p

Gradent The slope of the tangent at a point to a curve is called the gradient of the curve at that point.

The gradient of the curve y = f(x) at P is \(\left(\frac{d y}{d x}\right)_{P}\)

Note:

• If \(\left(\frac{d y}{d x}\right)_{P}\) = o then the tangent to the curve at P is parallel to the x – axis. The tangent, in this case, is called a horizontal tangent.
• If \(\left(\frac{d y}{d x}\right)_{P}\) = +∞ or -∞ i.e., if \(\left(\frac{d y}{d x}\right)_{P}\) = 0 then the tangent to the curve at P is perpendicular to x – axis. The tangent, in this case, is called a vertical tangent.
• If \(\left(\frac{d y}{d x}\right)_{P}\) does not exist then there exists no tangent to the curve at P.

Equation of Tangent:
The equation of the tangent at the point P(x₁, y₁) to the curve y = f(x) is y – y₁ = m(x – x₁)
where m = \(\left(\frac{d y}{d x}\right)_{P}\)

Note:

• x – intercept of the tangent = x₁ – \(\frac{y_{1}}{m}\) = x₁ – y₁m
• y – intercept of the tangent = y₁ – mx₁ = y₁ – x₁m

Normal To A Curve
Let P be a point in the curve y = f(x). The line passing through P and perpendicular to the tangent at P to the curve is called the normal to the curve at P.

The slope of the normal to the curve y = f(x) at P is m’ = \(\frac{1}{m}=-\left(\frac{d x}{d y}\right)_{P}\) where m = \(\left(\frac{d y}{d x}\right)_{P}\) ≠ 0

The equation of the normal at p(x₁, y₁) to the curve y = f(x) is
y – y₁ = \(\frac{-1}{m}\)(x – x₁) where m = \(\left(\frac{d y}{d x}\right)_{P}\)
i.e., y – y₁ = \(\left(\frac{d y}{d x}\right)_{P}\)(x – x₁)

Infinitesimals:
Let x be a finite variable quantity and be a minute change in x. Such a quanitity , which is very small when compared to x and which is smaller than any pre-assigned small quantity, is called an infinitesimal or an infinitesimal of first order. If δX is an infinitesimal then (δX)² , (δX)³ , …….. are called infinitesimals respectively of 2nd order, 3rd order….

If A is a finite quantity and is an infinitesimal then A. δX , A. (δX)² , A. (δX)³, ………. are also infinitesimals and they are infinitesimals respectively of first order, second order, third order
Definition: A quantity α = α(x) is called an infinitesimal as x → a if \({Lt}_{x \rightarrow a} α(x)\) = 0

Theorem:
Let y = f (x) be a differentiable function at x and be a small change in x. Then f'(x) and \(\frac{δ}{δ x}\) differ by an infinitesimal G(δx) as δx → 0 , where δy = f (x + δx) – f (x).

Differential
Definition: If y = f (x) is a differentiable function of x then f ’(x)S is called the differential of f. It is denoted by df or dy.
dy = f'(x)δx or df = f'(x)δx.
Note: δf = df i.e., error in f is approximately equal to differential of f

Approximations:
We have δf = f (x + δx) – f (x) ………………(1)
⇒ df ≅ f (x + δx) – f (x)
⇒ f’ (x)δx ≅ f (x + δx) – f (x)
⇒ f (x + δx) ≅ f (x) + f’ (x)δx
If we know the value of f at a point x, then the approximate value of f at a very nearby point x + δX can be calculated with the help of above formula.

Errors
Definition: Let y=f(x) be a function defined in a nbd of a point x. Let δx be a small change in x and δy be the corresponding change in y.
If δx is considered as an error in x, then

• δf is called the absolute error or error in y,
• \(\frac{δy}{y}\) is called the relative error (or proportionate error) in y,
  \(\frac{δy}{y}\) × 100 is called the percentage error in y corresponding to the error ox in x.

Lengths Of Tangent, Normal, Subtangent And Sub Normal
Definition :
Let y = f (x) be a differentiable curve and P be a point on the curve.
Let the tangent and normal at P to the curve meet the x – axis in T and N respectively. Let M be the projection of P on the x – axis. Then
(i) PT is called the length of the tangent,
(ii) PN is called the length of the normal
(iii) TM is called the length of the subtangent,
(iv) MN is called and length of the subnormal at the point P.

Let P(x₁, y₁) be a point on the curve y = f (x). Then

• The length of the tangent to the curve at P is \(\left|\frac{y_{1}}{m} \sqrt{1+m^{2}}\right|\)
• the length of the normal to the curve at P is |y₁\(\sqrt{1+m^{2}}\)|
• the length of the subtangent to the curve at P is \(\left|\frac{y_{1}}{m}\right|\)
  the length of the subnormal to the curve at P is |y₁m| where m = \(\left(\frac{d y}{d x}\right)_{P}\)

Angle between two curves:
If two curves intersect at P then the angle between the tangents to the curves at P is called the angle between the curves at P.

Angle between the curves:
Let y = f (x) and y = g(x) be two differentiable curves intersecting at a point P. Let m₁ = [ f'(x)]_P , m₂ = [g'(x)]_P be the slopes of the tangents to the curves at P. If θ is the acute angle between the curves at P then tan θ = \(\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|\)

Note:

• If m₁ = m₂ then θ = 0. In this case the two curves touch each other at P. Hence the curves have a common tangent and a common normal at P.
• If m₁m₂ = -1then θ = \(\frac{\pi}{2}\). In this case the curves cut each other orthogonally at P.
• If m₁ = 0 and \(\frac{1}{m_{2}}\) = 0 then the tangents to the curves are parallel to the coordinate axes.
• Therefore the angle between the curves is θ = \(\frac{\pi}{2}\)

Rate of Changes:
Let y = f(x) be defined on an interval (a,b). let δy be change in y corresponding to a change δx in x.
Then , \(\frac{\Delta y}{\Delta x}\) is called average rate of change of y. if \(\underset{δx \rightarrow 0}{L t} \frac{\Delta y}{\Delta x}\) exists finitely, the this limit (i.e., \(\frac{d y}{d x}\)) is called rate of change of y with respect to x .

Note:
\(\left(\frac{d y}{d x}\right)_{a t \mathrm{x}=\mathrm{c}}\) represents the rate of change of y with respect to x at x = c.

Rectilinear Motion:
Let a particle start at a point O on a line L and move along the line. After a time of t units, let the particle be at P and OP = s. Since s is dependent on time t, we write s=s(t). s is called the displacement of the particle during time t.
The rate of change of displacement s is called the velocity of the particle and it is denoted by v = \(\frac{d s}{d t}\).
The rate of change of velocity is called acceleration. It is denoted by a.
a = \(\frac{d v}{d t}=\frac{d}{d t}\left(\frac{d s}{d t}\right)=\frac{d^{2} s}{d t^{2}}\)

Angular Velocity:
At time t, let P be the position of a moving point, Q be the position of the point after an interval δt. Let ∠xop = 6 and ∠POQ = δθ

The angular velocity of P at O is\(\underset{x \rightarrow 0}{L t} \frac{δθ}{δt}=\frac{dθ}{d t}\)
Angular velocity w\(\frac{d \theta}{d t}\)
The angular acceleration of P at O is \(\frac{d \omega}{d t}=\frac{d}{d t}\left(\frac{d \theta}{d t}\right)=\frac{d^{2} \theta}{d t^{2}}\)

Maxima And Minima
Monotonic Functions Over An Internal
Definition: A function f :[a, b] → R is said to be
(i) Monotonically increasing (or non – decreasing) on [a, b] if x₁ < x₂ ⇒ f (x₁) ≤ f (x₂) ∀ x₁,x₂ ∈ [a, b]
(ii) Monotonically decreasing (or non – increasing) on [a, b] if x₁ < x₂ ⇒ (x₁) ≥ f (x₂) ∀ x₁,x₂ ∈ [a, b]
(iii) strictly increasing on [a, b] if x₁ < x₂ ⇒ (x₁) < f (x₂) ∀ x₁,x₂ ∈ [a, b]
(iv) strictly decreasing on [a, b] if x₁ < x₂ ⇒ (x₁) > f (x₂) ∀ x₁,x₂ ∈ [a, b]
(v) a monotonic function on [a, b] if f is either monotonically increasing or monotonically decreasing on [a, b].

Theorem:
Let f be a function defined in a nbd of a point a and f be differentiable at a. Then

• f'(a) > 0 ⇒ f is locally increasing at a.
• f >(a) < 0 ⇒ f is locally decreasing at a. Note: Let f be continuous on [a, b] and differentiable on (a, b). Then (i) f'(x) ≥ 0 ∀ x ∈ (a,b) ⇒ f (x) is increasing on [a, b]. (ii) f'(x) ≤ 0 ∀ x ∈ (a,b) ⇒ f (x) is decreasing on [a, b].
• f'(x) > 0 ∀ x ∈ (a,b) ⇒ f (x) is strictly increasing on [a, b].
• f'(x) < 0 ∀ x ∈ (a,b) ⇒ f (x) is strictly decreasing on [a, b].
• f'(x) = 0 ∀ x ∈ (a,b) ⇒ f (x) is a constant function on [a,b].

Greatest And Least Values
Definition: Let f be a function defined on a set A and l ∈ f (A). Then l is said to be
(i) the maximum value or the greatest value of f in A if f (x) ≤ l ∀ x ∈ A .
(ii) the minimum value or the least value of f in A if f (x ) ≥ l ∀ x ∈ A

Local Maximum And Local Minimum Values
Let f be a function defined in a nbd of a point ‘a’ then f is said to have
(i)a local maximum (value) or a relative maximum at a if ∃ a δ > 0 such that f (x) < f (a) ∀ x ∈ (a – δ, a) ∪ (a, a + δ) . In this case a is called a point of local maximum of f and f (a) is its local maximum value.

(ii) a local minimum (value) or relative minimum at a if 3 a 5> 0 such that f (x) > f (a) ∀ x ∈ (a – δ, a) ∪ (a, a + δ). In this case a is called a point of local minimum and f (a) is its local minimum value.

Theorem:
Let f be a differentiable function in a nbd of a point a. The necessary condition for f to have local maximum or local minimum at a is f ‘(a) = 0.

First Derivative Test
Let f be a differentiable function in a nbd of a point a and f'(a) = 0. Then

• f (x) has a relative maximum at x =a if ∃ a δ > 0 such that
  x ∈ (a – δ, a) ⇒ f'(x) > 0 and x ∈ (a, a + δ) ⇒ f'(x) < 0 .
• f(x) has a relative minimum at x= a if 3 a 8> 0 such that
  x ∈ (a – δ, a) ⇒ f'(x) < 0 and x ∈ (a, a +δ) ⇒ f'(x) < 0 .
• f (x) has neither a relative maximum nor a minimum at x=a if f'(x) has the same sign for all x ∈ (a – δ, a) ∪ (a, a + δ) .

Second Derivative Test
Let f (x) be a differentiable function in a nbd of a point ‘a’ and let f”(a) exist.

• If f'(a) = 0 and f”(a) < 0 then f (x) has a relative maximum at f (x) and the maximum value at a is f (a).
• If f'(a) = 0 and f”(a) > 0 then f (x) has a relative minimum at f (x) and the minimum value at a is f (a).

Absolute Maxima And Absolute Minima
Let f be a function defined on [a, b]. Then

• Absolute maximum of f on [a, b] = Max.{ f (a), f (b) and all relative maximum values of f in (a, b)}.
• Absolute minimum of f on [a, b] = Min.{ f (a), f (b) and all relative minimum values of f in (a,b)}.

Note :

• maximum and minimum values are called extrimities.
• If f'(a) = 0 , then f is said to be stationary at a and f(a) is called the stationary value of f. and (a, f(a)) is called a stationary point of f.

Mean Value Theorems
Rolle’sTheorem :If a function f : [a, b] → R is such that

• It is continuous on [a, b]
• It is derivable on (a, b) and i
• f(a) = f(b) then there exists at least one C ∈ (a,b) such that f ‘(C) = 0.

Lagrange’s mean -value theorem or first mean – value theorem :
If a function f : [a, b] → R is such that

• It is continuous on [a, b].
• It is derivable on (a, b) then there exists at least one C ∈ (a,b) such that \(\frac{f(b)-f(a)}{b-a}\) = f ‘(C)

Use these Inter 1st Year Maths 1B Formulas PDF Chapter 9 Differentiation to solve questions creatively.

Intermediate 1st Year Maths 1B Differentiation Formulas

→ f'(x) = \({Lt}_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

→ Δy = f(x + Δx) – f(x)

→ \(\frac{d}{d x}\) (u ± v) = \(\frac{d u}{d x} \pm \frac{d v}{d x}\)

→ \(\frac{d}{d x}\)(uv) = u\(\frac{d v}{d x}\) + v\(\frac{d u}{d x}\)

→ \(\frac{d}{d x}\)(uvw) = uv\(\frac{d w}{d x}\) + uw\(\frac{d v}{d x}\) + vw\(\frac{d u}{d x}\)

→ \(\frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}\)

→ \(\frac{d}{d x}\)(xⁿ) = nx^n-1

→ \(\frac{d}{d x}\)(x) = 1

→ \(\frac{d}{d x}\)(sin x) = cos x

→ \(\frac{d}{d x}\)(cos x) = -sin x

→ \(\frac{d}{d x}\)(tan x) = sec²x

→ \(\frac{d}{d x}\)(cot x) = -cosec²x

→ \(\frac{d}{d x}\)(sec x) = sec x tan x

→ \(\frac{d}{d x}\)(cosec x) = -cosec x cot x

→ \(\frac{d}{d x}\)(sin^-1x) = \(\frac{1}{\sqrt{1-x^{2}}}\)

→ \(\frac{d}{d x}\)(cos^-1x) = –\(\frac{1}{\sqrt{1-x^{2}}}\)

→ \(\frac{d}{d x}\)(tan^-1x) = \(\frac{1}{1+x^{2}}\)

→ \(\frac{d}{d x}\)(cot^-1x) = –\(\frac{1}{1+x^{2}}\)

→ \(\frac{d}{d x}\)(sec^-1x) = \(\frac{1}{|x| \sqrt{x^{2}-1}}\)

→ \(\frac{d}{d x}\)(cosec^-1x) = \(\frac{-1}{|x| \sqrt{x^{2}-1}}\)

→ \(\frac{d}{d x}\)(e^x) = e^x

→ \(\frac{d}{d x}\)(a^x) = a^xlog a

→ \(\frac{d}{d x}\)(log x) = \(\frac{1}{x}\)

→ \(\frac{d}{d x}\)|x| = \(\frac{|x|}{x}\)

→ \(\frac{d}{d x}\)(log |x|) = \(\frac{1}{x}\)

→ \(\frac{d}{d x}\)(x^x) = x^x(1 + log x)

→ \(\frac{d}{d x}\)(sinh x) = cosh x

→ \(\frac{d}{d x}\)(cosh x) = sinh x

→ \(\frac{d}{d x}\)(tanh x) = sech²x

→ \(\frac{d}{d x}\)(coth x) = -cosech²x

→ \(\frac{d}{d x}\)(sech x) = -sech x tanh x

→ \(\frac{d}{d x}\)(cosech x) =-cosech x coth x

→ \(\frac{d}{d x}\)(sinh^-1x) = \(\frac{1}{\sqrt{1+x^{2}}}\)

→ \(\frac{d}{d x}\)(cosh^-1x) = \(\frac{1}{\sqrt{x^{2}-1}}\)

→ \(\frac{d}{d x}\)(tanh^-1x) = \(\frac{1}{1-x^{2}}\)

→ \(\frac{d}{d x}\)coth^-1x) = \(\frac{1}{1-x^{2}}\)

→ \(\frac{d}{d x}\)(sech^-1x) = \(\frac{-1}{|x| \sqrt{1-x^{2}}}\)

→ \(\frac{d}{d x}\)(cosech^-1x) = \(\frac{-1}{|x| \sqrt{x^{2}+1}}\)

→ \(\frac{\mathrm{d} y}{\mathrm{dx}}=\frac{\mathrm{d} y}{\mathrm{du}} \cdot \frac{\mathrm{d} u}{\mathrm{dx}}\)

→ \(\frac{d}{d x}\left(\frac{1}{x^{n}}\right)=\frac{-n}{x^{n+1}}\)

Parametric Differentiation
If x =g (t) and y =h (t) are two functions of t then y may be expressed as a function of x, say y = f(x), by eliminating the real variable t. Such a variable t is called a parameter. Also, x and y are said to be defined parametrically. The equations x =g (t), y =h (t) are called the parametric equations of y = f(x).

The process of finding \(\frac{d y}{d x}\) when x and y are defined parametrically, is called parametric differentiation.

Theorem:
Let x = g(t) and y = h(t) be two differentiable functions of t such that g^-1 exists.
Then \(\frac{d y}{d x}=\left(\frac{d y}{d t}\right) /\left(\frac{d x}{d t}\right)\)

Implicit Differentiation
Let y be an implicit function of x defined by a relation of the type f(x, y) = 0. Then the method of finding the derivative \(\frac{d y}{d x}\) of such an implicitly defined function is called implicit differentiation.
In this method, we differentiate f(x, y) = 0 w.r.t x (treating y as a function of x) to get an equation of the type F(x,y, \(\frac{d y}{d x}\)) = 0.
By solving this equation we find \(\frac{d y}{d x}\).

Logarithmic Differentiation
If a given function is of the form y = f(x) g(x) where f(x) and g(x) are differentiable functions of x, we apply logarithms to transform it into a suitable form so as to find . This method of finding \(\frac{d y}{d x}\) is called logarithmic differentiation.
This method is also useful in finding the derivatives of functions which are products of a number of functions.

Theorem:
If y = f(x) g(x) where f(x) and g(x) are differentiable functions then \(\frac{d y}{d x}\) = f (x)^g(x)[\(\frac{g(x)}{f(x)}\)f'(x) + g'(x)log f(x)]
Proof:
y = f(x)g(x), f(x) > 0
logy = g(x) log f(x)

Differentiating both sides w.r.t x,

Method of Substitution:
In the case of inverse trigonometric functions with typical arguments direct differentiation of the function to find the derivative may become quite tedious and cumbersome. In such cases the derivative can be easily found by using proper trigonometric substitutions and transformations. The application of this method is illustrated in the following examples.

Derivative Of One Function W.R.T Another Function
Let f(x) and g(x) be two differentiable functions with a common domain of differentiability then the derivative of u = f(x) w.r.t v = g(x) is given by
\(\frac{d u}{d v}=\left(\frac{d u}{d x}\right) /\left(\frac{d v}{d x}\right)=\frac{f^{\prime}(x)}{g^{\prime}(x)}\)

Second Order Derivatives:
If y = f(x) is a differentiable function of x then its derivative f ‘(x) is a function of x. If f ‘(x) is a differentiable function then its derivative is called the second order derivative of f(x). It is denoted by f ”(x) and f”(x) = \({Lt}_{h \rightarrow 0} \frac{f^{\prime}(x+h)-f^{\prime}(x)}{h}\)
f ”(x) is also denoted by f”(x) = \({Lt}_{h \rightarrow 0} \frac{f^{\prime}(x+h)-f^{\prime}(x)}{h}\) or \(\frac{d^{2} f}{d x^{2}}\) or D²y or y” or y₂.

Derivatives of Inverse Trignometric Functions
If y = sin^-1x, x ∈ (-1, 1) then \(\frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}\)
Proof:
y = sin^-1x ⇒ x = sin y

Theorem:
y = cos^-1x, x ∈(-1, 1) then \(\frac{d y}{d x}=\frac{-1}{\sqrt{1-x^{2}}}\)

Theorem:
y = Tan^-1x, x ∈(-1, 1) then \(\frac{d y}{d x}=\frac{1}{1+x^{2}}\)
Proof:
y = Tan^-1x ⇒ x = tan y where x ∈ R and y ∈ \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)
x = tan y ⇒ \(\frac{d y}{d x}\) = sec²y = 1 + tan²y = 1 + x² > 0
∴ \(\frac{d y}{d x}=\left(\frac{d x}{d y}\right)^{-1}=\frac{1}{1+x^{2}}\), x ∈ R

Theorem:
y = Cot^-1x, x ∈ R then \(\frac{d y}{d x}=\frac{-1}{1+x^{2}}\)

Theorem:
y = Sec^-1x, x ∈ (-∞, -1) ∪ (1, ∞) then \(\frac{d y}{d x}=\frac{1}{|x| \sqrt{x^{2}-1}}\)

Theorem:
\(\frac{d}{d x}\)(Cosec^-1x) = \(\frac{-1}{|x| \sqrt{x^{2}-1}}\), |x| > 1

Derivatives of Hyperbolic Functions

1. \(\frac{d}{d x}\)(sinh x) = cosh x, x ∈ R
\(\frac{d}{d x}\)(sinh x) = \(\frac{d}{d x}\left(\frac{e^{x}-e^{-x}}{2}\right)\)
= \(\frac{1}{2}\)[\(\frac{d}{d x}\)(e^x) – \(\frac{d}{d x}\)(e^-x] = \(\frac{1}{2}\)[e^x + e^-x]
= cosh x, exists for all x.

2. \(\frac{d}{d x}\)(cosh x) = sinh x, x ∈ R
\(\frac{d}{d x}\)(cosh x) = \(\frac{d}{d x}\left(\frac{e^{x}+e^{-x}}{2}\right)\)
= \(\frac{1}{2}\)[\(\frac{d}{d x}\)(e^x) + \(\frac{d}{d x}\)(e^-x] = \(\frac{1}{2}\)[e^x – e^-x] = sinh x, ∀x

3. \(\frac{d}{d x}\)(tanh x) = sech²x, x ∈ R

4. \(\frac{d}{d x}\)(Coth x) = -cosec²x, x ∈ R

= \(\frac{-1}{\sinh ^{2} x}\) = -cosech²x exists for all x ≠ 0

Derivatives of Inverse Hyperbolic Functions:
1. \(\frac{d}{d x}\)(sinh^-1x) = \(\frac{1}{\sqrt{1+x^{2}}}\), x ∈ R
Let y = sinh^-1x, x, y ∈ R
Then x = sinh y, sinh y is differentiable for all y ∈ R
\(\frac{d x}{d y}\) = cosh y and cosh y > 0 ∀ x ∈ R
\(\frac{d y}{d x}=\left(\frac{d x}{d y}\right)^{-1}=\frac{1}{\cosh y}=\frac{1}{\sqrt{1+\sinh ^{2} y}} .=\frac{1}{\sqrt{1+x^{2}}}\) exists
\(\frac{d}{d x}\)(sinh^-1x) = \(\frac{1}{\sqrt{1+x^{2}}}\), x ∈ R

2. \(\frac{d}{d x}\)(cosh^-1x) = \(\frac{1}{\sqrt{x^{2}-1}}\), x ∈ (1, ∞)

3. \(\frac{d}{d x}\)(tanh^-1x) = \(\frac{1}{1-x^{2}}\), |x| < 1
Let y = Tanh^-1x, x ∈ (-1, 1), y ∈ R
Then x = tanhy. Tanh y is differetniable on R
∴ \(\frac{d x}{d y}\) = sech² ⇒ \(\frac{d y}{d x}=\frac{1}{{sech}^{2} y}=\frac{1}{1-\tanh ^{2} y}=\frac{1}{1-x^{2}}\) exists for all x ∈ (-1, 1)

4. \(\frac{d}{d x}\)(coth^-1x) = \(\frac{1}{1-x^{2}}\), |x| > 1

5. \(\frac{d}{d x}\)(sech^-1x) = \(\frac{-1}{x \sqrt{1-x^{2}}}\), x ∈ (0, 1)

6. \(\frac{d}{d x}\)(cosech^-1x) = \(\frac{-1}{x \sqrt{1+x^{2}}}\), x ≠ 0
Let y = cosech^-1x, x ∈ R – {0}, y ∈ R- {0}.
Then x = cosech y, cosech y is differentiable on R – {0}.

Partial Differentiation
Let u = f (x, y) be a function of two independent varaibles x and y.
(i) If \({Lt}_{h \rightarrow 0} \frac{f(x+h, y)-f(x, y)}{h}\) exists then the limit is called the partial derivatie of u with respect to x. It is denoted by \(\frac{\partial u}{\partial x}\) or u_x or \(\frac{\partial f}{\partial x}\) or f_x.

(ii) If \({Lt}_{k \rightarrow 0} \frac{f(x, y+k)-f(x, y)}{k}\) exists then the limit is called the partial derivative of u with respect to x. It is denoted by \(\frac{\partial u}{\partial y}\) or u_x or \(\frac{\partial f}{\partial y}\) or f_x.
∴ \(\frac{\partial u}{\partial x}={Lt}_{h \rightarrow 0} \frac{f(x+h, y)-f(x, y)}{h}\) and \(\frac{\partial u}{\partial y}=\underset{k \rightarrow 0}{L t} \frac{f(x, y+k)-f(x, y)}{k}\)

Note :

• The partial derivative of u w.r.t. x is the ordinary derivative of u w.r.t. x treating the other variable y (and its functions) as constant
• The partial derivative of u w.r.t. y is the ordinary derivative of u w.r.t. y treating the other variable x (and its functions) as constant.

Differentiation of Composite Functions
1. If V = g(U) and U = f(x, y) then (i) \(\frac{\partial V}{\partial x}=\frac{d V}{d U} \cdot \frac{\partial U}{\partial x}\) (ii) \(\frac{\partial V}{\partial y}=\frac{d V}{d U} \cdot \frac{\partial U}{\partial y}\)

2. If Z = f(x, y) and x = g(t); y = h(t) then \(\frac{d Z}{d t}=\frac{\partial Z}{\partial x} \cdot \frac{d x}{d t}+\frac{\partial Z}{\partial y} \frac{d y}{d t}\) is called the total differential coefficient of Z w.r.t. t.

3. If f(x, y) = c where c is constant, then \(\frac{d y}{d x}=-\frac{\partial f}{\partial x} / \frac{\partial f}{\partial y}\).

Partial Derivatives Of Second Order:
Definition : If U = f(x, y) then \(\frac{\partial \mathbf{U}}{\partial \mathrm{x}}, \frac{\partial \mathbf{U}}{\partial \mathbf{y}}\) are called the partial derivatives of first order and they are functions of x, y. The partial derivatives of \(\frac{\partial U}{\partial x}\) and \(\frac{\partial U}{\partial y}\), if they exist, are called the second order partial derivatives . They are denoted by

Homogeneous Functions:
A function u = f (x, y) is said to be a homogeneous function of degree n in the variables in x and y if f(kx, ky) = kⁿf(x, y) for all k or f (x, y) = xⁿ f\(\left(\frac{y}{x}\right)\) or f (x, y) = yⁿ f\(\left(\frac{y}{x}\right)\).

Euler’S Theorem:
If u = f (x, y) is a homogeneous function of degree n in the variables x,y then x\(\frac{\partial \boldsymbol{U}}{\partial x}\) + y \(\frac{\partial \boldsymbol{U}}{\partial y}\) = nU
Proof:
Since u = f (x, y) is a homogeneous function of degree n, we have U = xng (y / x) where g (y / x) is function of y/x.
\(\frac{\partial \boldsymbol{U}}{\partial x}\) = xⁿ.g'(y / x)\(\left(\frac{-y}{x^{2}}\right)\) + nx^n-1 .g (y / x) ………(1)
and \(\frac{\partial \boldsymbol{U}}{\partial y}\) = xⁿ .g'(y / x) \(\left(\frac{1}{x}\right)\)
x\(\frac{\partial \boldsymbol{U}}{\partial x}\) + y\(\frac{\partial \boldsymbol{U}}{\partial y}\) = x(x\(\left(\frac{-y}{x^{2}}\right)\)) + nxⁿg(y/x) + yx^n-1.g’9y/x)
= n.xⁿ.g (y / x) == nU.

Note :
If U = f (x, y, z) is a homogeneous function of degree n in x,y,z then
x\(\frac{\partial U}{\partial x}\) + y\(\frac{\partial U}{\partial y}\) + z\(\frac{\partial U}{\partial z}\) = nU

Theorem:
If U = f (x, y) is a homogeneous function of degree n in x,y then x²\(\frac{\partial^{2} U}{\partial x^{2}}\) + 2xy\(\frac{\partial^{2} U}{\partial x \partial y}\) + y²\(\frac{\partial^{2} U}{\partial y^{2}}\) = n(n – 1)U .
Proof:
Since U = f (x, y) is a homogeneous function of degree n, by Euler’s theorem, we have

Use these Inter 1st Year Maths 1B Formulas PDF Chapter 8 Limits and Continuity to solve questions creatively.

Intermediate 1st Year Maths 1B Limits and Continuity Formulas

Right Limit :
Suppose f is defined on (a, b) and Z ∈ R. Given ε < 0 , there exists δ > 0 such that a < x < a + δ => |f(x) – l| < ε, then l is said to be the right limit of’ f’ at ‘a’.
It is denoted by \({Lt}_{x \rightarrow a+} f(x)\)f(x) = l

Left Limit :
Suppose ‘ f’ is defined on (a, b) and Z e R. Given e > 0, there exists δ > 0 such that a – δ < x < a ⇒ |f(x) – l| < ε, then l is said to be the left limit of’ f’ at ’a’ and is denoted by \({Lt}_{x \rightarrow a-} \dot{f(x)}\) = l

Suppose f is defined in a deleted neighbourhood of a and l ∈ R
\({Lt}_{x \rightarrow a} f(x)=l \Leftrightarrow \underset{x \rightarrow a+}{L t} f(x)={Lt}_{x \rightarrow a-} \quad f(x)=l\)

Standard limits:

• \({Lt}_{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}\) = na^n-1
• \({Li}_{x \rightarrow 0} \frac{\sin x}{x}\) = 1 (x is in radians)
• \({Lt}_{x \rightarrow 0} \frac{\tan x}{x}\) = 1
• \({lt}_{x \rightarrow 0}(1+x)^{1 / x}\) = e
• \({Lit}_{x \rightarrow 0}\left(1+\frac{1}{x}\right)^{x}\) = e
• \({Lt}_{x \rightarrow 0}\left(\frac{a^{x}-1}{x}\right)\) = log_ea
• \({Lt}_{x \rightarrow a} \frac{x^{n}-a^{n}}{x^{m}-a^{m}}=\frac{n}{m}\)a^{n – m}
• \({Lt}_{x \rightarrow 0} \frac{\sin a x}{x}\) = a(x is in radians)
• \({Lt}_{x \rightarrow 0} \frac{\tan a x}{x}\) = a(x is in radians)
• \({Lt}_{x \rightarrow \infty}\left(1+\frac{p}{x}\right)^{Q x}\) = e^PQ
• \({lt}_{x \rightarrow 0} \frac{e^{x}-1}{x}\) = 1
• \({Lt}_{x \rightarrow 0}(1+p x)^{\frac{Q}{x}}\) = e^PQ

Intervals
Definition:
Let a, b ∈ R and a < b. Then the set {x ∈ R: a ≤ x ≤ b} is called a closed interval. It is denoted by [a, b]. Thus
Closed interval [a, b] = {x ∈ R: a ≤ x ≤ b}. It is geometrically represented by

Open interval (a,b) = {x ∈ R: a < x < b} It is geometrically represented by

Left open interval
(a, b] = {x ∈ R: a < x ≤ b}. It is geometrically represented by

Right open interval
[a, b) = {x ∈ R: a ≤ x < b}. It is geometrically represented by

[a, ∞) = {x ∈ R : x ≥ a} = {x ∈ R : a ≤ x < ∞} It is geometrically represented by

(a, ∞) = {x ∈ R : x > a} = {x ∈ R : a < x < ∞}

(-∞, a] = {x ∈ R : x ≤ a} = {xe R : -∞ < x < a}

Neighbourhood of A Point: Definition: Let ae R. If δ > 0 then the open interval (a – δ, a + δ) is called the neighbourhood (δ – nbd) of the point a. It is denoted by N_δ (a) . a is called the centre and δ is called the radius of the neighbourhood .
∴ N_δ(a) = (a – δ, a + δ) = {x ∈ R: a – δ< x < a + δ} = {x ∈ R: |x – a| < δ}

The set N_δ(a) – {a} is called a deleted δ – neighbourhood of the point a.
∴ N_δ(a) – {a} = (a – δ, a) ∪ (a, a + δ) = {x ∈ R :0 < | x – a | < δ}
Note: (a – δ, a) is called left δ -neighbourhood, (a, a + δ) is called right δ – neighbourhood of a

Graph of A Function:

Mod function:
The function f: R-R defined by f(x) = |x| is called the mod function or modulus function or absolute value function.
Dom f R. Range f [0, )

Reciprocal function :
The function f: R – {0} – R defined by
f(x) = \(\frac{1}{x}\) is called the reciprocal function,
Dom f = R – {0}= Range f = R

Identity function:
The function f R-R defined by f(x) = x is called the identity
It is denoted by I(x)

Limit of A Function
Concept of limit:
Before giving the formal definition of limit consider the following example.
Let f be a function defined by f (x) = \(\frac{x^{2}-4}{x-2}\). clearly, f is not defined at x = 2.
When x ≠ 2, x – 2 ≠ 0 and f(x) = \(\frac{(x-2)(x+2)}{x-2}\) = x + 2

Now consider the values of f(x) when x ≠ 2, but very very close to 2 and <2.

It is clear from the above table that as x approaches 2 i.e.,x → 2 through the values less than 2, the value of f(x) approaches 4 i.e., f(x) → 4. We will express this fact by saying that left hand limit of f(x) as x → -2 exists and is equal to 4 and in symbols we shall write \({lt}_{x \rightarrow 2^{-}} f(x)\) = 4

Again we consider the values of f(x) when x ≠ 2, but is very-very close to 2 and x > 2.

It is clear from the above table that as x approaches 2 i.e.,x → 2 through the values greater than 2, the value of f(x) approaches 4 i.e., f(x) → 4. We will express this fact by saying that right hand
limit of f(x) as x → 2 exists and is equal to 4 and in symbols we shall write \({lt}_{x \rightarrow 2^{+}} f(x)\) = 4

Thus we see that f(x) is not defined at x = 2 but its left hand and right hand limits as x → 2 exist and are equal.
When \({lt}_{\mathrm{x} \rightarrow \mathrm{a}^{+}} \mathrm{f}(\mathrm{x}), \mathrm{lt}_{\mathrm{x} \rightarrow \mathrm{a}^{-}}^{\mathrm{f}(\mathrm{x})}\) are equal to the same number l, we say that \(\begin{array}{ll}
l_{x \rightarrow a} & f(x)
\end{array}\) exist and equal to 1.

Thus, in above example,

One Sided Limits Definition Of Left Hand Limit:
Let f be a function defined on (a – h, a), h > 0. A number l1 is said to be the left hand limit (LHL) or left limit (LL) of f at a if to each
ε >0, ∃ a δ >0 such that, a – δ < x < a ⇒ |f (x) – l₁| < ε.
In this case we write \(\underset{x \rightarrow a-}{L t} f(x)\) = l₁ (or) \({Lt}_{x \rightarrow a-0} f(x)\) = l₁

Definition of Right Limit:
Let f be a function defined on (a, a + h), h > 0. A number l ₂is said to the right hand limit (RHL) or right limit (RL) of f at a if to each ε >0, ∃ a δ >0 such that, a – δ < x < a ⇒ |f (x) – l₂| < ε.
In this case we write \(\underset{x \rightarrow a+}{L t} f(x)\) = l₂ (or) \({Lt}_{x \rightarrow a+0} f(x)\) = l₂

Definition of Limit:
Let A ∈ R, a be a limit point of A and
f : A → R. A real number l is said to be the limit of f at a if to each ε > 0, ∃ a δ > 0 such that x ∈ A, 0 < |x – a| < δ ⇒ | f(x) – l| < ε. In this case we write f (x) → 2 (or) \({Lt}_{x \rightarrow a} f(x)\) = l. Note: 1.If a function f is defined on (a – h, a) for some h > 0 and is not defined on (a, a + h) and if \(\underset{x \rightarrow a-}{{Lt}} f(x)\) exists then \({Lt}_{x \rightarrow a} f(x)={Lt}_{x \rightarrow a-} f(x)\).
2. If a function f is defined on (a, a + h) for some h > 0 and is not defined on (a – h, a) and if \(\underset{x \rightarrow a+}{{Lt}} f(x)\) exists then \({Lt}_{x \rightarrow a} f(x)={Lt}_{x \rightarrow a+} f(x)\).

Theorem:
If \({Lt}_{x \rightarrow a} f(x)\) exists then \({Lt}_{x \rightarrow a} f(x)={Lt}_{x \rightarrow 0} f(x+a)={Lt}_{x \rightarrow 0} f(a-x)\)

Theorems on Limits WithOut Proofs
1. If f : R → R defined by f(x) = c, a constant then \({Lt}_{x \rightarrow a} f(x)\) = c for any a ∈ R.

2. If f: R → R defined by f(x) = x, then \({Lt}_{x \rightarrow a} f(x)\) = a i.e., \({Lt}_{x \rightarrow a} f(x)\) = a (a ∈ R)

3. Algebra of limits

vii) If f(x) ≤ g(x) in some deleted neighbourhood of a, then \({Lt}_{x \rightarrow a} f(x) \leq {Lt}_{x \rightarrow a} g(x)\)
viii) If f(x) ≤ h(x) < g(x) in a deleted nbd of a and \({Lt}_{x \rightarrow a} f(x)\) = l = \({Lt}_{x \rightarrow a} g(x)\) then \({Lt}_{x \rightarrow a} h(x)\) = l
ix) If \({Lt}_{x \rightarrow a} f(x)\) = 0 and g(x) is a bounded function in a deleted nbd of a then \({Lt}_{x \rightarrow a}\) f(x) g(x) = 0.

Theorem
If n is a positive integer then \({Lt}_{x \rightarrow a} x^{n}\) = aⁿ, a ∈ R

Theorem
If f(x) is a polynomial function, then \({Lt}_{x \rightarrow a}\)f(x) = f (a)

Evaluation of Limits:
Evaluation of limits involving algebraic functions.
To evaluate the limits involving algebraic functions we use the following methods:

• Direct substitution method
• Factorisation method
• Rationalisation method
• Application of the standard limits.

1. Direct substitution method:
This method can be used in the following cases:

• If f(x) is a polynomial function, then \({Lt}_{x \rightarrow a}\)f(x) = f (a).
• If f (x) = \(\frac{P(a)}{Q(a)}\) where P(x) and Q(x) are polynomial functions then \({Lt}_{x \rightarrow a}\)f(x) = \(\frac{P(a)}{Q(a)}\) provided Q(a) ≠ 0.

2. Factiorisation Method:
This method is used when \({Lt}_{x \rightarrow a}\)f (x) is taking the indeterminate form of the type 0 by the substitution of x = a.
In such a case the numerator (Nr.) and the denominator (Dr.) are factorized and the common factor (x – a) is cancelled. After eliminating the common factor the substitution x = a gives the limit, if it exists.

3. Rationalisation Method : This method is used when \({Lt}_{x \rightarrow a} \frac{f(x)}{g(x)}\) is a \(\frac{0}{0}\) form and either the Nr. or Dr. consists of expressions involving radical signs.

4. Application of the standard limits.
In order to evaluate the given limits , we reduce the given limits into standard limits form and then we apply the standard limits.

Theorem 1.
If n is a rational number and a > 0 then \({Lt}_{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}\) = n.a^n-1

Note:

• If n is a positive integer, then for any a ∈ R. \({Lt}_{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}\) = n.a^n-1
• If n is a real number and a> 0 then \({Lt}_{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}\) = n.a^n-1
• If m and n are any real numbers and a > 0, then \({Lt}_{x \rightarrow a} \frac{x^{m}-a^{m}}{x^{n}-a^{n}}=\frac{m}{n}\)a^m-n

Theorem 2.
If 0 < x < \(\frac{\pi}{2}\) then sin x < x < tan x.

Corollary 1:
If – \(\frac{\pi}{2}\) < x < 0 then tan x < x < sin x

Corollary 2:
If 0 < |x| < \(\frac{\pi}{2}\) then |sin x| < |x| < |tan x|

Standard Limits:

• \({Lt}_{x \rightarrow 0} \frac{\sin x}{x}\) = 1
• \(\underset{x \rightarrow 0}{L t} \frac{\tan x}{x}\) = 1
• \({Lt}_{x \rightarrow 0} \frac{e^{x}-1}{x}\) = 1
• \({Lt}_{x \rightarrow 0} \frac{a^{x}-1}{x}\) = log_ea
• \({Lt}_{x \rightarrow 0}(1+x)^{\frac{1}{x}}\) = e
• \({Lt}_{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}\) = e
• \({Lt}_{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}\) = e

Limits At Infinity
Definition:
Let f(x) be a function defined on A = (K,).
(i) A real number l is said to be the limit of f(x) at ∞ if to each δ > 0, ∃ , an M > 0 (however large M may be) such that x ∈ A and x > M ⇒ |f (x) – l|< δ.
In this case we write f (x) → l as x → +∞ or \({Lt}_{x \rightarrow \infty} f(x)\) = l.

(ii) A real number l is said to be the limit of f(x) at -∞ if to each δ > 0, ∃ > 0, an M > 0 (however large it may be) such that x ∈ A and x < – M ⇒ |f (x) – l| < δ. In this case, we write f (x) → l as x → -∞ or \({Lt}_{x \rightarrow \infty} f(x)\) = l.

Infinite Limits Definition:
(i) Let f be a function defined is a deleted neighbourhood of D of a. (i) The limit of f at a is said to be ∞ if to each M > 0 (however large it may be) a δ > 0 such that x ∈ D, 0 < |x – a| < δ ⇒ f(x) > M. In this case we write f(x) as x → a or \({Lt}_{x \rightarrow a} f(x)\) = +∞

(ii) The limit of f(x) at a is said be -∞ if to each M > 0 (however large it may be) a δ > 0 such that x ∈ D, 0 < | x – a | < δ ⇒ f(x) < – M. In thise case we write f(x) → -∞ as x → a or \({Lt}_{x \rightarrow a} f(x)\) = -∞

Indeterminate Forms:
While evaluation limits of functions, we often get forms of the type \(\frac{0}{0}, \frac{\infty}{\infty}\), 0 × -∞, 0⁰, 1^∞, ∞⁰ which are termed as indeterminate forms.

Continuity At A Point:
Let f be a function defined in a neighbourhood of a point a. Then f is said to be continuous at the point a if and only if \({Lt}_{x \rightarrow a} f(x)\) = f (a).
In other words, f is continuous at a iff the limit of f at a is equal to the value of f at a.

Note:

• If f is not continuous at a it is said to be discontinuous at a, and a is called a point of discontinuity of f.
• Let f be a function defined in a nbd of a point a. Then f is said to be
  • Left continuous at a iff \({Lt}_{x \rightarrow a-} f(x)\) = f (a).
  • Right continuous at a iff \({Lt}_{x \rightarrow a+} f(x)\) = f (a).
• f is continuous at a iff f is both left continuous and right continuous at a
  i.e, \({Lt}_{x \rightarrow a} f(x)=f(a) \Leftrightarrow {Lt}_{x \rightarrow a^{-}} f(x)=f(a)={Lt}_{x \rightarrow a+} f(x)\)

Continuity of A Function Over An Interval:
A function f defined on (a, b) is said to be continuous (a,b) if it is continuous at everypoint of (a, b) i.e., if \({Lt}_{x \rightarrow c} f(x)\) =f (c) ∀c ∈ (a, b)
II) A function f defined on [a, b] is said to be continuous on [a, b] if

• f is continuous on (a, b) i.e., \({Lt}_{x \rightarrow c} f(x)\) = f (c) ∀c ∈(a, b)
• f is right continuous at a i.e., \({Lt}_{x \rightarrow a+} f(x)\) = f (a)
• f is left continuous at b i.e., \({Lt}_{x \rightarrow b-} f(x)\) = f (b).

Note :

• Let the functions f and g be continuous at a and k€R. Then f + g, f – g, kf , kf + lg, f.g are continuous at a and \(\frac{f}{g}\) is continuous at a provided g(a) ≠ 0.
• All trigonometric functions, Inverse trigonometric functions, hyperbolic functions and inverse hyperbolic functions are continuous in their domains of definition.
• A constant function is continuous on R
• The identity function is continuous on R.
• Every polynomial function is continuous on R.

Use these Inter 1st Year Maths 1B Formulas PDF Chapter 7 The Plane to solve questions creatively.

Intermediate 1st Year Maths 1B The Plane Formulas

Definition: If F(x, y, z) is a polynomial of degree one, then the surface represented by F(x, y, z).= 0 is called a first degree

→ If A, B, C are three non-collinear points in space, then there exists exactly only one plane through A,

→ If a plane contains two points A and B, then the plane also contains the line \(\overleftrightarrow{A B}\).

→ If two planes intersect, then their intersection is a line.

→ If a plane π and a line L are perpendicular at a point P, then π contains every line that passes through P and perpendicular to L.

→ Equation of a plane in the normal form is lx + my + nz = p. Hence (l, m, n) are the D.C’s of the normal to the plane and p( > 0) is the perpendicular distance of the plane from the origin.

→ If (a, b, c) & (0,0,0), then the equation ax + by + cz + d = 0 is called the general form of the equation of a plane.

→ Equation of the plane n which contains the point A(x₀, y₀, z₀) and perpendicular to the line L with direction ratios (a, b, c) is a(x – xQ) + b(y – y0) + c(z – z ) = 0.

→ Equation of the plane passing through three non-collinear point A(x₁, y₁; z₁), B(x₂, y₂, z₂) and C(x₃ y₃, z₃) is \(\left|\begin{array}{ccc}
x-x_{1} & y-y_{1} & z-z_{1} \\
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{i}
\end{array}\right|\) = 0

→ Equation of the plane in the intercept form is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1.

→ The angle between the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is cos^-1 \(\frac{\left|a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}\right|}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\)

→ The perpendicular distance of the plance ax + by + cz + d = 0 from the point P(x₀ , y₀ , z₀) is \(\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

→ The distance between the parallel planes ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0 is \(\frac{\left|d_{1} \cdot d_{2}\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

Planes:
Definition:
A surface in space is said to be a plane surface or a plane if all the points of the straight line joining any two points of the surface lie on the surface.

Theorem:
The equation of the plane passing through a point (x₁, y₁, z₁) and perpendicular to a line whose direction ratios are a, b, c is a( x – x₁)+b( y – y₁)+c( z – z₁) = 0

Theorem:
The equation of the plane passing through a point (x₁, y₁, z₁) is a( x – x₁)+b( y – y₁)+c (z – z₁) = 0 where a,b,c are constants.

Theorem:
The equation of the plane containing three points (x₁, y₁, z₁), (x₂, y₂, z₂), (x₃, y₃, z₃) is \(\left|\begin{array}{ccc}
x-x_{1} & y-y_{1} & z-z_{1} \\
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1}
\end{array}\right|\) = 0

Normal Form of A Plane:
Theorem:
The equation of the plane which is at a distance of p from the origin and whose normal has the direction cosines (l, m, n) is lx + my + nz = p (or) x cos α+ y cos β + z cos γ = p.

Note:
Equation of the plane through the origin is lx + my + nz = 0
Note: The equation of the plane ax +by+cz+d=0 in the normal form is

where Σa² = a² + b² + c²

Perpendicular Distance From A Point to A Plane:
The perpendicular distance from the origin to the plane ax + by + cz + d = 0 is \(\frac{|d|}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

Theorem
The perpendicular distance from P(x1, y1, z1) to the plane ax + by + cz + d = 0 is \(\frac{\left|a x_{1}+b y_{1}+c z_{1}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

Theorem
Intercept form of the plane
The equation of the plane having a,b,c as x, y, z- intercepts respectively is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1

Theorem
The intercepts of the plane ax + by + cz + d = 0 are respectively \(\frac{-d}{a}, \frac{-d}{b}, \frac{-d}{c}\)

Angle Between Two Planes:
Definition: The angle between the normals to two planes is called the angle between the planes.

Theorem:
If θ is the angle between the planes a₁x + b₁ y + c₁z + d₁ = 0, a₂x + b₂y + c₂z + d₂ = 0 then cos θ = \(\frac{a_{1} a_{2}+b_{1} b_{1}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\)

Note:

• If θ is acute then cos θ = \(\frac{a_{1} a_{2}+b_{1} b_{1}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \mid\)
• The planes a₁x + b₁ y + c₁z + d₁ = 0, a₂x + b₂y + c₂z + d₂ = 0 are
  • parallel iff \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
  • Perpendicular iff . a₁a₂ + b₁b₂ + c₁c₂ = 0
• The given planes are perpendicular
  ⇔ θ = 90° ⇔ cos θ = 0 ⇒ a₁a₂ + b₁b₂ + c₁c₂ = 0

Theorem:
The equation of the plane parallel to the plane ax + by + cz + d = 0 is ax + by + cz + k = 0 where k is a constant.

Theorem:
The distance between the parallel planes ax + by + cz + d₁ = 0, ax + by + cz + d₂ = 0 is \(\frac{\left|d_{1}-d_{2}\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

Use these Inter 1st Year Maths 1B Formulas PDF Chapter 6 Direction Cosines and Direction Ratios to solve questions creatively.

Intermediate 1st Year Maths 1B Direction Cosines and Direction Ratios Formulas

Formulae and Synopsis :
If a line PQ makes angles α, β, γ with the co – ordinate axes, then cos α, cos β, cos γ are called
direction cosines of the line PQ. We take
l = cos α, m = cos β and n = cos γ
Relation between direction cosines is l² + m² + n² = 1

Direction Ratios :
An ordered triple of numbers proportional to the direction cosines of a line are defined as ‘Direction ratios’ of that fine.

→ If \(\frac{1}{a}=\frac{m}{b}=\frac{n}{c}\) then a, b, c are called direction ratios of the line.

→ If a, b, c are the direction ratios of a line, then its direction cosines are \(\left(\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, \frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, \frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}\right)\)

→ Direction ratios of the line joining P(x₁, y₁, z₁) and Q (x₂, y₂, z₂) are (x₂ – x₁, y₂ – y₁, z₂ – z₁) or (x₁ – x₂, y₁ – y₂, z₁ – z₂)

→ D.cs of the above line \(\frac{x_{2}-x_{1}}{P Q}, \frac{y_{2}-y_{1}}{P Q}, \frac{z_{2}-z_{1}}{P Q}\) or \(\frac{x_{1}-x_{2}}{P Q}, \frac{y_{1}-y_{2}}{P Q}, \frac{z_{1}-z_{2}}{P Q}\)

→ Angle between the lines whose D.cs are (l₁ m₁, n₁) and (l₂, m₂, n₂) is given by cos θ = l₁l₂ + m₁m₂ + n₁n₂

→ If these lines are perpendicular, then l₁l₂ + m₁m₂ + n₁n₂ = 0

Angle Between Two Lines:
The angle between two skew lines is the angle between two lines drawn parallel to them through any point in space.

Direction Cosines:
If α, β, γ are the angles made by a directed line segment with the positive directions of the coordinate axes respectively, then cos α, cos β, cos γ are called the direction cosines of the given line and they are denoted by l, m, n respectively Thus l = cos α, m = cos β, n = cos γ

The direction cosines of \(\overline{o p}\) are l = cos α, m = cos β, n = cos γ.
If l, m, n are the d.c’s of a line L is one direction then the d.c’s of the same line in the opposite direction are -l, -m, -n.

Note :

• The angles α, β, γ are known as the direction angles and satisfy the condition 0 < α, β, γ < π.
• The sum of the angles α, β, γ is not equal to 2β because they do not lie in the same plane.
• Direction cosines of coordinate axes.

The direction cosines of the x-axis are cos0, cos\(\frac{\pi}{2}\), cos\(\frac{\pi}{2}\) i.e., 1, 0, 0
Similarly the direction cosines of the y-axis are (0, 1, 0) and z-axis are (0, 0, 1)

Theorem:
If P(x, y, z) is any point in space such that OP = r and if l, m, n are direction cosines of \(\overline{O P}\) then x = lr,y = mr, z = nr.
Note: If P(x, y, z) is any point in space such that OP = r then the direction cosines of \(\overline{O P}\) are
Note: If P is any point in space such that OP =r and direction cosines of OP are l,m,n then the point P =(lr,mr,nr)
Note: If P(x,y,z) is any point in space then the direction cosines of OP are \(\frac{x}{\sqrt{x^{2+} y^{2}+z^{2}}}, \frac{y}{\sqrt{x^{2+} y^{2}+z^{2}}}, \frac{z}{\sqrt{x^{2+} y^{2}+z^{2}}}\)

Theorem:
If l, m, n are the direction cosines of a line L then l² + m² + n² = 1.
Proof:

l = cos α = \(\frac{x}{r}\), m = cos β = \(\frac{y}{r}\), n = cos γ = \(\frac{z}{r}\) ⇒ cos²α + cos²β + cos²γ = \(\frac{x^{2}}{r^{2}}+\frac{y^{2}}{r^{2}}+\frac{z^{2}}{r^{2}}\)
= \(\frac{x^{2}+y^{2}+z^{2}}{r^{2}}=\frac{r^{2}}{r^{2}}\)
∴ l² + m² + n² = 1

Theorem:
The direction cosines of the line joining the points P(x₁, y₁, z₁),Q(x₂, y₂, z₂) are
\(\left(\frac{x_{2}-x_{1}}{r}, \frac{y_{2}-y_{1}}{r}, \frac{z_{2}-z_{1}}{r}\right)\) where r = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}\)

Direction Ratios:
A set of three numbers a,b,c which are proportional to the direction cosines l,m,n respectively are called DIRECTION RATIOS (d.r’s) of a line.
Note : If (a, b, c) are the direction ratios of a line then for any non-zero real number λ , (λa, λb, λc) are also the direction ratios of the same line.
Direction cosines of a line in terms of its direction ratios
If (a, b, c) are direction ratios of a line then the direction cosines of the line are ±\(\left(\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, \frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, \frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}\right)\)

Theorem:
The direction ratios of the line joining the points are (x₂ – x₁, y₂ – y₁, z₂ – z₁)

Angle Between Two Lines:
If (l₁, m₁, n₁) and (l₂, m₂, n₂) are the direction cosines of two lines θ and is the acute angle between them, then cos θ = |l₁l₂ + m₁m₂ + n₁n₂|

Note.
If θ is the angle between two lines having d.c’s (l₁, m₁, n₁) and (l₂, m₂, n₂) then
sin θ = \(\sqrt{\sum\left(l_{1} m_{2}-l_{2} m_{1}\right)^{2}}\) and tan θ = \(\frac{\sqrt{\sum\left(l_{1} m_{2}-l_{2} m_{1}\right)^{2}}}{\left|l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}\right|}\) where θ ≠ \(\frac{\pi}{2}\)

Note :
The condition for the lines to be perpendicular is l₁l₂ + m₁m₂ + n₁n₂ = 0
The condition for the lines to be parallel is \(\frac{l_{1}}{l_{2}}=\frac{m_{1}}{m_{2}}=\frac{n_{1}}{n_{2}}\)

Theorem:
If (a₁, b₁, c₁) and (a₂, b₂, c₂) are direction ratios of two lines and θ is the angle
between them then cos θ = \(\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\)

Note:

• If the two lines are perpendicular then a₁a₂ + b₁b₂ + c₁c₂ = 0
• If the two lines are parallel then \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
• If one of the angle between the two lines is 0 then other angle is 180° – θ

Use these Inter 1st Year Maths 1B Formulas PDF Chapter 5 Three Dimensional Coordinates to solve questions creatively.

Intermediate 1st Year Maths 1B Three Dimensional Coordinates Formulas

→ Distance from the origin to P(x, y, z) = OP = \(\sqrt{x^{2}+y^{2}+z^{2}}\)

→ Distance between the points P(x₁, y₁ z₁) and Q (x₂, y₂, z₂) is

→ PQ = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}+\left(z_{1}-z_{2}\right)^{2}}\)

→ Translation of axes : If the origin is shifted to the point (h, k, l) under translation of axes, then x = x’ + h, y = y’ + k, z = z’ + l where (x, y, z) are the old co-ordinates and (x’, y’, z’) are the new co-ordinates.

Coordinates of a Point In Space (3-D):
Let \(\overline{X^{\prime} O X}, \overline{Y^{\prime} O Y}\) and \(\overline{Z^{\prime} O Z}\) be three mutually perpendicular straight lines in space, intersecting at O. This point O is called origin.

Axes :
The three fixed straight lines \(\overline{X^{\prime} O X}, \overline{Y^{\prime} O Y}\) and \(\overline{Z^{\prime} O Z}\) are respectively called X-axis ,Y-axis and Z-axis. The three lines taken together are called rectangular coordinate axes.

Coordinate Planes:
The plane containing the axes of Y and Z is called yz-plane. Thus yoz is the yz plane. Similarly the plane zox containing the axes of z and x is called ZX-plane and the plane xoy is called the xy-plane and contains x axis and y axis.
The above three planes are together called the rectangular co ordinate plane.

Octants:
The three co ordinate planes divide the whole space into 8 parts called octants.

Coordinates of A Point:
Let P be any point in the space. Draw through P , three planes parallel to the three co ordinate planes meeting the axes of X,Y,Z in the points A,B and C respectively. Then if OA = x, OB = y and OC = z, the three numbers x,y,z taken in this order are called the co ordinates of the point P and we refer the point as (x,y,z) . Any one of these x,y,z will be positive of negative according as it is measured from O along the corresponding axis, in the positive or negative direction.
Another method of finding coordinates of a point.
The coordinates x,y,z of a point P are the perpendicular distances of P from the three co ordinate planes YZ,ZX and XY respectively.

From fig PN = z, PL = x and PM = y
Therefore point P =(x,y,z)
Note. On YZ- plane, a point has x coordinate as zero and similarly on zx-plane y coordinates and on xy-plane z coordinates are zero.

For any point on the

• X-axis, Y,Z coordinates are equal to zero,
• Y-axis, X,Z coordinates are equal to zero,
• Zaxis, X,Y coordinates are equal to zero.

Distance Between The Points:
The distance between the points and is given by PQ = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}\)

Note:
If is the origin and is a point in space, then
OP = \(\sqrt{(x-0)^{2}+(y-0)^{2}+(z-0)^{2}}=\sqrt{x^{2}+y^{2}+z^{2}}\)

Section Formula:
P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) be two points in space and let R be a point on the line segment joining P and Q such that it divides \(\overline{P Q}\) internally in the ratio m: n . Then the coordinate of are \(\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}, \frac{m z_{2}+n z_{1}}{m+n}\right)\)

(ii) P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) be two points in space and let R be a point on the line segment joining P and Q such that it divides \(\overline{P Q}\) externally in the ratio m: n . Then the coordinate of are \(\left(\frac{m x_{2}-n x_{1}}{m-n}, \frac{m y_{2}-n y_{1}}{m-n}, \frac{m z_{2}-n z_{1}}{m-n}\right)\), m ≠ n

Mid Point:
The mid point of the linesegment joining the points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) is
\(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right)\)

Centroid of a triangle:
The coordinates of the centriod of the triangle with vertices A(x₁, y₁, z₁) B(x₂, y₂, z₂) and C(x₃, y₃, z₃) is \(\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}+\frac{z_{1}+z_{2}+z_{3}}{3}\right)\)

Centroid of a tetrahydron:
The coordinates of the centriod of the tetrahedron with vertices A(x₁, y₁, z₁) B(x₂, y₂, z₂), C(x₃, y₃, z₃) and D(x₄, y₄, z₄) is \(\left(\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}, \frac{y_{1}+y_{2}+y_{3}+y_{4}}{4}, \frac{z_{1}+z_{2}+z_{3}+z_{4}}{4}\right)\)

Translation of Axes :

Let P(x,y,z) and A(h,k,l) be two points is space w.r.t the frame of reference OXYZ. Now treating A as the origin, let \(\overline{A X^{1}}, \overline{A Y^{1}}, \overline{A Z^{1}}\) be the new axes parallel to
\(\overline{O X}, \overline{O Y}, \overline{O Z}\) respectively. If (x’, y’, z’) are the coordinates of P w.r.t
AX¹Y¹Z¹ then x¹ = x – h, y¹ = y – k, z¹ = z – l.

Example: Origin is shifted to the point (1,2 -3). Find the new coordinates of (1,0,-1)
Answer:
(x, y, z) = (1,0,-1) and( h, k, l) = (1,2 -3).
Now new coordinates are X = x – h = 1 -1 = 0
Y = y – k = 0 – 2 = -2
Z = z – l = -1 + 3 = 2
herefore new coordinates are ( 0,-2, 2)