Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B The Straight Line Solutions Exercise 3(d) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1B The Straight Line Solutions Exercise 3(d)

I. Find the angle between the following straight lines.

Question 1.
y = 4 – 2x, y = 3x + 7
Solution:
y = 4 – 2x ⇒ 2x + y – 4 = 0
3x – y + 7 = 0

Question 2.
3x + 5y = 7, 2x – y + 4 = 0
Solution:

Question 3.
y = – √3 + 5, y = $$\frac{1}{\sqrt{3}} x-\frac{2}{\sqrt{3}}$$
Solution:
m1 = √3, m2= $$\frac{1}{\sqrt{3}}$$
m1m2 = (-√3).$$\frac{1}{\sqrt{3}}$$ = 1
The lines are perpendicular θ = $$\frac{\pi}{2}$$

Question 4.
ax + by = a + b, a(x-y) + b(x + y) = 2b
Solution:
ax + by = a + b, (a + b) x + (- a + b) y = 2b

Find the length of the perpendicular drawn from the point given against the following straight lines.

Question 5.
5x – 2y + 4 = 0, (-2, -3)
Solution:
Length of the perpendicular

Question 6.
3x – 4y + 10 = 0, (3, 4)
Solution:
Length of the perpendicular
$$=\frac{|3.3-4.4+10|}{\sqrt{9+16}}=\frac{3}{5}$$

Question 7.
x – 3y – 4 = 0, (0,0)
Solution:
Length of the perpendicular
$$=\frac{|0-0-4|}{\sqrt{1+9}}=\frac{4}{\sqrt{10}}$$

Find the distance between the following parallel lines.

Question 8.
3x – 4y = 12, 3x – 4y = 7
Solution:
Equation of the lines are
3x – 4y – 12 = 0
3x – 4y – 7 = 0

Distance between parallel lines
$$=\frac{\left|c_{1}-c_{2}\right|}{\sqrt{a^{2}+b^{2}}}=\frac{|-| 2+7 \mid}{\sqrt{9+16}}=\frac{5}{5}=1$$

Question 9.
5x – 3y – 4 = 0, 10x – 6y – 9 = 0
Solution:
Equations of the lines can be taken as
10x – 6y – 8 = 0
10x – 6y – 9 = 0

Distance between parallel lines
$$=\frac{|-8+9|}{\sqrt{100+36}}=\frac{1}{2 \sqrt{34}}$$

Question 10.
Find the equation of the straight line parallel to the line 2x + 3y 4- 7 = 0 and passing through the point (5, 4).
Solution:
Equation of the given line is 2x + 3y + 7 = 0
The required line is parallel to this line
Equation of the parallel line is 2x + 3y = k
This line passes through p (5, 4)
10 + 12 = k ⇒ k = 22
Equation of the required line is
2x + 3y = 22
2x + 3y — 22 = 0

Question 11.
Find the equation of the straight line per-pendicular to the line 5x – 3y + 1 = 0 and passing through the point (4, – 3).
Solution:
Equation of the given line is 5x – 3y + 1 = 0
Equation of the perpendicular line is 3x + 5y + k = 0
This line passes through P (4, -3)
12 – 15 + k = 0 ⇒ k = 3
Equation of the required line is
3x + 5y + 3 = 0

Question 12.
Find the value of k, if the straight lines 6x – 10y + 3 = 0 and kx – 5y + 8 = 0 are parallel.
Solution:
Equations of the given lines are
6x – 10y + 3 = 0
kx – 5y + 8 = 0
These lines are parallel
a1b2 = a2b1
-30 = -10 k
k = 3

Question 13.
Find the value of p, if the straight lines 3x 4- 7y – 1 = 0 and 7x – py + 3 = 0 are mutually perpendicular.
Solution:
Equations of the given lines are
3x + 7y – 1 = 0
7x – py + 3 = 0
These lines are perpendicular
⇒ a1a2 + b1b2 = 0
3.7 + 7(- p) = 0
7p = 21 ⇒ p = 3

Question 14.
Find the value of k, if the straight lines y – 3kx + 4 = 0 and (2k-1) x-(8k-1) y – 6 = 0 are perpendicular.
Solution:
Equations of the given lines are
-3kx + y + 4 = 0
(2k- 1)x- (8k- 1)y- 6 = 0
These lines are perpendicular
⇒ -3k(2k – 1) – 1(8k – 1) = 0
– 6k² +.3k – 8k +1 = 0
6k² + 5k – 1 = 0
(k + 1) (6k – 1) = 0
k = -1 or 1/6

Question 15.
(- 4,5) is a vertex of a square and one of its diagonal is 7x – y + 8 = 0. Find the equation of the other diagonal.

Solution:
ABCD is a square Equation of the diagonal AC is 7x – y + 8 = 0
The other diagonal BD is perpendicular to AC.
Equation of BD can be taken asx + 7y + k = 0
BD passes through D (- 4, 5)
-4 + 35 + k = 0 ⇒ k = 4-35 = -31
Equation of BD is x + 7y- 31 = 0

II.

Question 1.
Find the equations of the straight lines passing through (1, 3) and i) parallel to ii) perpendicular to the line passing through the points (3, -5) and (-6,1).

Solution:
A (3, -5), B(- 6, 1) are the given points
Slope of AB = $$\frac{-5-1}{3+6}=\frac{-6}{9}=\frac{-2}{3}$$

i) The required line is parallel to AB and passes through (1, 3)
Equation of the required line is
y – 3 = $$\frac{-2}{3}$$(x – 1)
3y – 9 = -2x + 2
2x + 3y – 11 = 0

ii) The required line AE is perpendicular to AD.
Equation of AE can be taken as 3x – 2y + k = 0
This line passes through A (1, 3)
3 – 6 + k = 0 ⇒ k = 6 – 3 = 3
Equation of AE is 3x – 2y + 3 = 0.

Question 2.
The line $$\frac{x}{a}-\frac{y}{b}$$ = 1 meets the X-axis at P. Find the equation of the line perpen-dicular to this line at P.
Solution:

Equation of PQ is $$\frac{x}{a}-\frac{y}{b}$$ = 1
Equation of X-axis is y = 0
$$\frac{x}{a}$$ = 1 ⇒ x = a
Co-ordinates of P are (a, 0)
PR is perpendicular to PQ
Equation of PR = $$\frac{x}{b}+\frac{y}{a}$$ = k
This line PR passes through P is (a, 0) a
$$\frac{a}{b}$$ + 0 = k ⇒ k = a/b
Equation of PR is $$\frac{x}{b}+\frac{y}{a}=\frac{a}{b}$$ = 1

Question 3.
Find the equation of the line perpendi-cular to the line 3x + 4y + 6 = 0 and making an intercept – 4 on X-axis.
Solution:
Equation of the given line is 3x + 4y + 6 = 0
Equation of the perpendicular line is
4x – 3y = k
$$\frac{4x}{k}-\frac{3y}{k}$$ = 1
$$\frac{x}{\left(\frac{k}{4}\right)}+\frac{y}{\left(-\frac{k}{3}\right)}=1$$
x – intercept = $$\frac{k}{4}$$ = – 4 ⇒ k = -16
Equation of the required line is 4x – 3y = -16
⇒ 4x – 3y + 16 = 0

Question 4.
A (-1,1), B (5, 3) are opposite vertices of a square in the XY plane. Find the equation of the other diagonal, (not passing through A, B) of the square.
Solution:
A (-1, 1), B (5, 3) are opposite vertices of the square.
Slope of AB $$\frac{1-3}{-1-5}=\frac{-2}{-6}=\frac{1}{3}$$

The other diagonal perpendicular to AB
Slope of CD = – $$\frac{1}{m}$$ = -3 m
‘O’ is the point of intersection of the diagonals
Co-ordinates of O are$$\left(\frac{-1+5}{2}. \frac{1+3}{2}\right)$$ = (2, 2)
CD passes through O (2, 2)
Equation of CD is y – 2 = -3 (x – 2)
= -3x + 6
3x + y – 8 = 0

Question 5.
Find the foot of the perpendicular drawn from (4, 1) upon the straight line 3x – 4y + 12 = 0.
Solution:
Equation of the line is 3x – 4y + 12 = 0
If (x2, y2) is the foot of the perpendicular from (x1, y1) on the line
ax + by + c = 0, then

Question 6.
Find the foot of the perpendicular drawn from (3, 0) upon the straight line 5x + 12y – 41 = 0.
Solution:
Equation of the line is 5x + 12y – 41 = 0
If (x2, y2) is the foot of the perpendicular from (x1, y1) on the line ax + by + c = 0,

Question 7.
x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining the points A, B. If A = (-1, -3), find the co-ordinates of B.

Solution:
If PQ is the perpendicular bisector of AB, then B is the image of A in the line PQ.
Equation of PQ is x – 3y – 5 = 0

Co-ordinates of A are (-1, -3)
If (x2, y2) is the image of (x1, y1) about the line ax + by + c = 0, then

Question 8.
Find the image of the point (1, 2) in the straight line 3x + 4y – 1 = 0,
Solution:
Equation of the line is 3x + 4y – 1 = 0 ‘
If (x2, y2) is the image of (x1, y1) in the line ax + by + c = 0, then

Question 9.
Show that the distance of the point (6, – 2) from the line 4x + 3y = 12 is half the distance of the point (3,4) from the line 4x – 3y = 12.
Solution:

Equation of AB is 4x + 3y – 12 = 0
PQ = Length of the perpendicular from
$$P=\frac{|24-6-12|}{\sqrt{16+9}}=\frac{6}{5}$$
Equation of CD is 4x – 3y – 12 = 0
RS = Length of the perpendicular from
$$R=\frac{|12-12-12|}{\sqrt{16+9}}=\frac{12}{5}$$
∴ PQ = $$\frac{1}{2}$$RS

Question 10.
Find the locus of the foot of the perpendi¬cular from the origin to a variable straight line which always passes through a fixed point (a, b).

Solution:
Suppose m is the slope of the line AB
Equation of AB is y – b = m (x – a)
= mx – ma
mx – y + (b – ma) = 0 ……………. (1)

OK is perpendicular to A B and passes through the origin O. Suppose co-ordinates of K are (x, y)
Equation of AB is x + my – 0 ……………. (2)

m = -x/y
Substituting in (1)
$$-\frac{x^{2}}{y}-y+b+\frac{x}{y} \cdot a=0$$.
– x² – y² + by + ax = 0
or x² + y² – ax – by = 0
Locus of K is x² + y² – ax – by = 0

III.

Question 1.
Show that the lines x – 7y – 22 = 0, 3x + 4y + 9 = 0 and 7x + y – 54 = 0 form a right angled isosceles triangle.
Solution:
Given lines x – 7y – 22 = 0 …………. (1)
3x + 4y + 9 = 0 …………. (2)
7x + y – 54 = 0 …………. (3)

Let ‘A’ be the angle between (1), (2)

Let B be the angle between (2), (3)

Let ‘C’ be the angle between (3), (1)

∴ Given lines form a right angled isosceles triangle.

Question 2.
Find the equation of the straight lines passing through the point (-3, 2) and making an angle of 45° with the straight line 3x – y + 4 = 0.
Solution:
Given point
P(x1, y1) = (-3, 2)
Given line 3x – y + 4 = 0 ………….. (1)
Slope = m = – $$\frac{a}{b}$$ = 3

m – 3 = -1 + 3m
2m = – 4 or m = -2
$$\frac{m-3}{1+3 m}$$ = -1 ⇒ m – 3 = -1 – 3m
4m = 2 ⇒ m = 1/2
Case (i) m = – 2
Equation of the line PQ is
y – 2 = -2(x+3)
= -2x – 6
2x + y + 4 = 0

Case (ii) m =$$\frac{1}{2}$$
Equation of the line PR is
y – 2 = $$\frac{1}{2}$$ (x + 3)
2y – 4 = x + 3
x – 2y + 7 = 0

Question 3.
Find the angles of the triangle whose sides are x + y- 4 = 0,2x + y- 6 = 0, 5x + 3y -15 = 0.
Solution:
Equation of AB is x + y – 4 = 0
Equation of BC is 2x + y – 6 = 0
Equation of AC is 5x + 3y – 15 = 0

Question 4.
Proye that the feet of the perpendiculars from the origin on the lines x + y = 4, x + 5y = 26 and 15x – 27y = 424 are collinear.
Solution:
Given lines
x + y – 4 = 0 ……… (1)
x + 5y – 26 = 0 ……… (2)
15x – 27y – 424 = 0 ……… (3)

Let P(x2, y2) be the foot of the ⊥le of (x1, y1) = (0, 0) on (1)

⇒ x2 – 0 = 2, y2 – 0 = 2
⇒ x2 = 2, y2 = 2
∴ P = (2, 2)
Let Q(x3, y3) be the foot of the ⊥le of (x1 y1) = (0, 0) on (2)

x3 = 1, y3 = 5 ⇒ Q = (1, 5)
Let R(x4, y4) be the foot of the ⊥le of (x1, y1) = (0, 0) on (3)

Equation of the line through P,Q is
$$\frac{x-2}{1-2}=\frac{y-2}{5-2}$$ ⇒ 3x + y – 8 = 0 ………. (4)
Substitute R (x4, y4) in (4)
⇒ 3$$\frac{1060}{159}$$ – 12 – 8
= 20 – 20 = 0
∴ Foot of perpendiculars of origin on the lines lies on a straight line.

Question 5.
Find the equations of the straight lines passing through the point of intersection of the lines 3x + 2y + 4 = 0,2x + 5y = 1 and whose distance from (2, -1) is 2.
Solution:
Equations of the lines passing through the point of intersection of the line
L1 ≡ 3x + 2y + 4 = 0, L3 ≡ 2x + 5y – 1 = 0 is
L1 + λL2 = 0
(3x + 2y + 4) + λ (2x + 5y- 1) = 0
(3 + 2λ)x + (2 + 5λ) y + (4 – λ) = 0 ………….. (1)
Given distance from (2, -1) to (1) = 2

⇒ (-λ + 4)2 = 9 + 4λ2 + 12λ + 4 + 25λ2 + 20λ
⇒ 28λ2 + 40λ – 3 = 0
⇒ 28λ2 – 2λ + 42λ – 3 = 0
⇒ (2λ + 3) (14λ – 1) = 0
⇒ λ = $$\frac{1}{14}$$, λ = – $$\frac{3}{2}$$
From (1)
If λ = $$\frac{1}{14}$$ ⇒ 4x + 3y + 5 = 0
If λ = – $$\frac{3}{2}$$
⇒ y – 1 = 0 are the required line equations.

Question 6.
Each side of a square is of length 4 units. The centre of the square is (3, 7) and one of its diagonals is parallel to y = x. Find the co-ordinates of its vertices.
Solution:
Let ABCD be the square.
Point of intersection of the diagonals is the centre P(3, 7)

From P draw PM⊥AB. Then M is midpoint ofAB
∴ AM = MB = PM = 2
Since a diagonal is parallel to y = x, its sides are parallel to the co-ordinate axes.
M(3, 5) ⇒ A(1, 5), B(5, 5), C(5, 9), D(1, 9)

Question 7.
If ab > 0, find the area of the rhombus enclosed by the four straight lines ax ± by ± c = 0.
Solution:
Equation of AB is ax + by + c = 0 ………….. (1)
Equation of CD is ax + by – c = 0 ………….. (2)
Equation of BC is ax – by + c = 0 ………….. (3)
Equation of AD is ax – by – c = 0 ………….. (4)

Solving (1) and (3), co-ordinates of B are (-$$\frac{c}{a}$$ 0)
Solving (1) and (4) co-ordinates of A are (0, –$$\frac{c}{b}$$
Solving (2) and (3) co-ordinates of C are (0, $$\frac{c}{b}$$
Solving (2) and (4) co-ordinates of D are ($$\frac{c}{a}$$, 0)
Area of rhombus ABCD = $$\frac{1}{2}$$|∑x1 (y2 – y4)|

Question 8.
Find the area of the parallelogram whose sides are 3x + 4y + 5 = 0, 3x + 4y – 2 = 0, 2x + 3y + 1 = 0 and 2x + 3y – 7 = 0
Solution:
Given sides are 3x + 4y + 5 = 0 ……….. (1)
3x + 4y – 2 = 0 ……….. (2)
2x + 3y + 1 = 0 ……….. (3)
2x + 3y – 7 = 0 ……….. (4)

Area of parallelogram formed by (1), (2), (3), (4)

Question 9.
A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find the equation of the path the he should follow.
Solution:
By solving
2x – 3y + 4 = 0
3 x + 4y – 5 = 0

Given line 6x – 7y + 8 = 0
Required line is 7x + 6y + k = 0 …….. (1)

From (1) 7x + 6y – $$\frac{125}{17}$$
119x + 102y – 125 = 0

Question 10.
A ray of light passing through the point (1, 2) reflects on the X – axis at a point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.
Solution:
Let m be the slope then equation of line passing through (1, 2).
y – 2 = m(x – 1)
$$\frac{y-2}{x-1}=m$$
Let -m be the slope the equation of line passing through (5, 3)
y – 3 = -m(x – 5)