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:

m_{1} = √3, m_{2}= \(\frac{1}{\sqrt{3}}\)

m_{1}m_{2} = (-√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

a_{1}b_{2} = a_{2}b_{1}

-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

⇒ a_{1}a_{2} + b_{1}b_{2} = 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 (x_{2}, y_{2}) is the foot of the perpendicular from (x_{1}, y_{1}) 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 (x_{2}, y_{2}) is the foot of the perpendicular from (x_{1}, y_{1}) 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 (x_{2}, y_{2}) is the image of (x_{1}, y_{1}) 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 (x_{2}, y_{2}) is the image of (x_{1}, y_{1}) 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(x_{1}, y_{1}) = (-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(x_{2}, y_{2}) be the foot of the ⊥^{le} of (x_{1}, y_{1}) = (0, 0) on (1)

⇒ x_{2} – 0 = 2, y_{2} – 0 = 2

⇒ x_{2} = 2, y_{2} = 2

∴ P = (2, 2)

Let Q(x_{3}, y_{3}) be the foot of the ⊥^{le} of (x_{1} y_{1}) = (0, 0) on (2)

x_{3} = 1, y_{3} = 5 ⇒ Q = (1, 5)

Let R(x_{4}, y_{4}) be the foot of the ⊥^{le} of (x_{1}, y_{1}) = (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 (x_{4}, y_{4}) 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

L_{1} ≡ 3x + 2y + 4 = 0, L_{3} ≡ 2x + 5y – 1 = 0 is

L_{1} + λL_{2} = 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 = \(\frac{1}{2}\)|∑x_{1} (y_{2} – y_{4})|

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)