Well-structured Class 11 ISC Maths S Chand Solutions Chapter 16 The Straight Line Ex 16(f) facilitate a deeper understanding of mathematical principles.

## S Chand Class 11 ICSE Maths Solutions Chapter 16 The Straight Line Ex 16(f)

Question 1.

Find the distance of the point P from the line AB in the following cases :

(i) P(4, 2), AB is 5x – 12y – 9 = 0,

(ii) P(0, 0), AB is h(x + h) + k (y + k) = 0.

Solution:

(i) ⊥ distance of point P(4, 2) from given line AB

(ii) ⊥ distance of point P(0, 0) from given line AB

Question 2.

Calculate the length of the perpendicular from (7, 0) to the straight line 5x + 12y – 9 = 0 and show that it is twice the length of the perpendicular from (2, 1).

Solution:

Given eqn. of straight line be

5x +12y – 9 = 0

required length of ⊥ from P(7, 0) to given

Clearly the length of ⊥ from P(7, 0) to given line is twice the length of ⊥ from Q (2, 1) to same given line.

Question 3.

The point A(0, 0), B(1, 7), C(5, 1) are the vertices of a triangle. Find the length of the perpendicular from A to BC and hence the area of the △ABC.

Solution:

eqn. of line BC is given by using two point form, be

Question 4.

Find the lengths of altitudes of the triangle whose sides are given by

3x – 4y = 5, 4x + 3y = 5 and x + y = 1.

Solution:

Given eqns. of sides are

3x – 4y = 5 …(1)

4x + 3y = 5 …(2)

and x + y = 1 …(3)

First of all, we find all the vertices of triangle whose sides are along given lines (1), (2) and (3).

We solve three eqns. in pairs to get the vertices of triangle.

On solving eqn. (1) and (2) simultaneously, we have

x = \(\frac { 7 }{ 5 }\); y =\(\frac { -1 }{ 5 }\)

On solving eqn. (2) and (3) simultaneously ; we have

y = – 1 and x = 2

On solving eqn. (1) and (3) simultaneously ; we have

x = \(\frac { 9 }{ 7 }\) and y = \(\frac { -2 }{ 7 }\)

∴ length of altitude from A\(\left(\frac{7}{5}, \frac{-1}{5}\right)\) to BC

Question 5.

If P is the perpendicular distance of the origin from the line whose intercepts on the axes are a and b, show that

\(\frac{1}{p^2}\) = \(\frac{1}{a^2}\) = \(\frac{1}{b^2}\)

Solution:

eqn. of line whose intercepts on the axes are a and b be given by

\(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 …(1)

p = length of ⊥ from (0, 0) to given eqn. (1)

Question 6.

Find the perpendicular distance between the lines

(i) 3x + 4y + 5 = 0, 3x + 4y + 17 = 0.

(ii) 9x + 40y – 20 = 0, 9x + 40y + 21 = 0.

(iii) y = mx + c, y = mx + d.

Solution:

(i) Given eqns. of lines are

3x + 4y + 5 = 0 …(1)

and 3x + 4y +17 =0 …(2)

Slope of both lines =-\(\frac{3}{4}\)

∴ lines (1) and (2) are parallel.

For any point on line (1), we put x = 0 in eqn. (1); we have y = –\(\frac{5}{4}\)

Thus any point on line (1) be \(\left(0,-\frac{5}{4}\right)\)

∴ required distance between parallel lines = ⊥ distance from \(\left(0,-\frac{5}{4}\right)\) to line (2)

(ii) Given lines are;

9x + 40y – 20 = 0 …(1)

9x + 40y – 21 = 0 …(2)

Slopes of lines (1) and (2) = –\(\frac{9}{40}\)

∴ both lines are parallel.

For any point on line (1); we put x = 0 in eqn. (1) ;

we have y = \(\frac{1}{2}\)

∴ any point on line (1) be \(\left(0, \frac{1}{2}\right) \text {. }\)

Thus required distance between parallel lines

= ⊥ distance of point \(\left(0, \frac{1}{2}\right)\) to line (2)

(iii) Equations of given lines are

y = mx + c …(1)

y = mx + d …(2)

Clearly both lines are parallel.

For any point on line (1) ;

we put x = 0 in eqn. (1); we have y = c

∴ any point on line (1) be (0, c).

∴ required distance between parallel lines = distance of point (0, c) from line (2)

Question 7.

Find the equations of two straight lines which are parallel to the straight line x + 7y + 2 = 0, and a unit distance from the point (2, – 1).

Solution:

eqn. of given straight line be

x + 7y + 2 = 0 …(1)

∴ eqn. of line parallel to line (1) be given by

x + 7y + k = 0 …(2)

given ⊥ distance from the point (2, – 1) to line (2) = 1

putting the value of k in eqn. (2); we have x + 7y + 5(1 ± √ 2) = 0 be the required eqn.’s of lines.

Question 8.

Find the equations of the two straight lines drawn through the point (0, 1) on which the perpendiculars dropped from the point (2, 2) are each of unit length.

Solution:

eqn. of straight line passing through the point (0, 1) and is having slope m be given by

y – 1 = m(x – 0) [using one point form]

⇒ mx – y + 1 = 0 …(1)

It is given that, length of ⊥ drawn from point (2, 2) to line (1) be of unit length.

On squaring both sides; we have

(2m – 1)^{2} = m^{2} + 1

⇒ 4m^{2} – 4m + 1 = m^{2} + 1

⇒ 3m^{2} – 4m = 0

⇒ m(3m – 4) = 0 ⇒ m = 0, \(\frac { 4 }{ 3 }\)

putting m = 0 in eqn. (1); we have

– y + 1 = 0 ⇒ y = 1

putting m = \(\frac { 4 }{ 3 }\) in eqn. (1); we have

\(\frac { 4x }{ 3 }\) – y + 1 = 0

⇒ 4x – 3y + 3 = 0

Question 9.

A straight line is parallel to the lines 3x – y – 3 = 0 and 3x – y + 5 = 0, and lies between them. Find its equation if its distances from these lines are in the ratio 3 : 5.

Solution:

Given eqns. of lines are

3x – y – 3 = 0 …(1)

3x – y + 5 = 0 …(2)

Let the eqn. of line which is mid-parallel to lines (1) and (2) be given by

3x – y + k = 0 …(3)

To find any point on line (3); we put x = 0 in eqn. (3) ; we have, y = k

∴ any point on line (3) be (0, k).

Also it is given that, distances of line (3) from given lines (1) and (2) are in the ratio 3 : 5

On squaring both sides; we have

\(\frac{(k+3)^2}{(5-k)^2}\) = \(\frac{9}{25}\)

⇒ 25 (k^{2} + 6k + 9) = 9 (k^{2} – 10k + 25)

⇒ 16 k^{2} + 240k = 0

⇒ 16 k (k + 15) = 0

⇒ k = 0, – 15

∴ from (3); 3x – y = 0

or 3x – y – 15 = 0

are the required eqns. of lines.

Question 10.

Find the equation of the locus of a point P which is equidistant from the st. line 3x – 4y + 2 = 0 and the origin.

Solution:

eqn. of given line be

3x – 4y + 2 = 0 …(1)

Let P(x, y) be any point whose locus is to be find out.

s.t distance of P(x, y) from O(0, 0) = length of ⊥ from P(x, y) to line (1)

on squaring both sides, we have

⇒ 25 (x^{2} + y^{2}) = 9x^{2} + 16y^{2} + 4 – 24xy – 16y + 12x

⇒ 16x^{2} + 9y^{2} + 24xy + 16y – 12x – 4 = 0

which is the required eqn. of locus.

Question 11.

A point P is such that the sum of the squares of its distances from the two axes of co-ordinates is equal to the square of its distance from the line x – y = 1. Find the equation of the locus of P.

Solution:

Eqn. of coordinate axes be given by x = 0 and y = 0

Let P(x, y) be any point whose locus is to be find out.

According to given condition, we have

Question 12.

Show that the equation to the parallel line mid-way between the parallel lines ax + by + c_{1} = 0 and ax + by + c_{2} = 0 is ax + by + \(\frac{c_1+c_2}{2}\) = 0.

Solution:

Eqns. of given lines re

ax + by + c_{1} = 0 …(1)

and ax + by + c_{2} = 0 …(2)

Let the eqn. of line which is mid-parallel to both given lines be

ax + by + c = 0 …(3)

For any point on line (3); we put x = 0 in eqn. (3); we have y =-\(\frac { c }{ b }\)

Thus any point on line (3) be \(\left(0,-\frac{c}{b}\right)\)

Since line (3) is mid-parallel between line (1) and line (2).

∴ ⊥ distance of point \(\left(0,-\frac{c}{b}\right)\) from line (1)

= ⊥ distance of point \(\left(0,-\frac{c}{b}\right)\) from line (2)

Question 13.

Prove that the line 12x – 5y – 3 = 0 is midparallel to the lines 12x – 5y + 7 = 0 and 12x – 5y – 13 =0 .

Solution:

Given eqns. of lines are

12x – 5y – 3 = 0 …(1)

12x – 5y + 7 = 0 …(2)

and 12x – 5y – 13 = 0 …(3)

Now line (1) is mid-parallel to lines (2) and (3)

If ⊥ distance of any point on line (1) from line (2) is equal to ⊥ distance of any point on line (1) from line (3).

For any point on line (1) ∴ we put x = 0 in eqn. (1) we have y = –\(\frac { 3 }{ 5 }\)

∴ Any point on line (1) be \(\left(0,-\frac{3}{5}\right)\)

⊥ distance of \(\left(0,-\frac{3}{5}\right)\) from line (2)