# Class 11 RD Sharma Solutions – Chapter 23 The Straight Lines- Exercise 23.19

### Question 1. Find the equation of a straight line through the point of intersection of the lines 4x – 3y = 0 and 2x – 5y + 3 = 0 and parallel to 4x + 5 y + 6 = 0.

**Solution:**

From the question we have,

The equations of lines which are :

4x – 3y = 0 ………….. (1)

2x – 5y + 3 = 0 …….. (2)

Now we find the equation of the straight line that pass through the

points of intersection of equations (1) and (2) is given,

4x − 3y + λ (2x − 5y + 3) = 0 ……………………… (3)

(4 + 2λ)x + (− 3 − 5λ)y + 3λ = 0……………….. (4)

y = (4 + 2λ / 3 + 5λ)x + 3λ/(3 + 5λ)

The line is parallel to 4x + 5y + 6 = 0 or, y = -4x/5 – 6/5

4 + 2λ / 3 + 5λ = -4/5

λ = -16/15

Now put the value of λ in eq(4), we get

(4 – (32/15))x – (3 – (80/15))y – 48/15 = 0

Hence, the equation of line is 28x + 35y – 48 = 0

### Question 2. Find the equation of a straight line passing through the point of intersection of x + 2y + 3 = 0 and 3x + 4y + 7 = 0 and perpendicular to the straight line x – y + 9 = 0.

**Solution:**

From the question we have,

The equations of lines which are :

x + 2y + 3 = 0 ….. (1)

3x + 4y + 7 = 0 …. (2)

Now we find the equation of the straight line that pass through the

points of intersection of equations (1) and (2) is given,

So, x + 2y + 3 + λ(3x + 4y + 7) = 0 ………….. (3)

(1 + 3λ)x + (2 + 4λ)y + 3 + 7λ = 0 …………. (4)

y = – ((1 + 3λ)/(2 – 4λ))x – ((3 + 7λ)/(2 + 4λ))

So, the slope of the line is -(1 + 3λ/2 + 4λ)

It is given that the line whose slope is -(1 + 3λ/2 + 4λ)

is perpendicular to the straight line x – y + 9 = 0

So, -(1 + 3λ/2 + 4λ) × 1 = 1

λ = 1

Now put the value of λ in eq(3), we get

x + 2y + 3 + 1(3x + 4y + 7) = 0

x + y + 2 = 0

Hence, the equation of the line is x + y + 2 = 0

### Question 3. Find the equation of the line passing through the point of intersection of 2x – 7y + 11 = 0 and x + 3y – 8 = 0 and is parallel to (i)** x – axis (ii) y – axis.**

**Solution:**

From the question we have,

The equations of lines which are :

2x – 7y + 11 = 0 ….. (1)

x + 3y – 8 = 0 ….. (2)

Now we find the equation of the straight line that pass through the

points of intersection of equations (1) and (2) is given,

2x − 7y + 11 + λ(x + 3y − 8) = 0

………. (3)(2 + λ)x + (− 7 + 3λ)y + 11 − 8λ = 0 ………. (4)

(i)x – axis2 + λ = 0

λ = -2

Now, we put the value of λ in equation (4), we get

0 + (− 7 − 6)y + 11 + 16 = 0

13y − 27 = 0

Hence, the equation of the required line is 13y − 27 = 0

(2)y – axis7 + 3λ = 0

λ = 7/3

Now, we put the value of λ in equation (4), we get

(2 + 7/3)x + 0 + 11 – 8(7/3) = 0

13x – 23 = 0

Hence, the equation of the required line is 13x – 23 = 0

### Question 4. Find the equation of the straight line passing through the point of intersection of 2x + 3y + 1 = 0 and 3x – 5y – 5 = 0 and equally inclined to the axes.

**Solution:**

From the question we have,

The equations of lines which are :

2x + 3y + 1 = 0 ….. (1)

3x – 5y – 5 = 0 ….. (2)

Now we find the equation of the straight line that pass through the

points of intersection of equations (1) and (2) is given,

2x + 3y + 1 + λ(3x − 5y − 5) = 0 …………….. (3)

(2 + 3λ)x + (3 − 5λ)y + 1 − 5λ = 0 …………… (4)

y = – [(2 + 3λ) / (3 – 5λ)] – [(1 – 5λ) / (3 – 5λ)]

It is given that the line is equally inclined to the axes.

So, the slope of the line is either 1 or − 1.

Hence,

– [(2 + 3λ) / (3 – 5λ)] = 1 and – [(2 + 3λ) / (3 – 5λ)] = -1

-2 – 3λ = 3 – 5λ and 2 + 3λ = 3 – 5λ

λ = 5/2 and 1/8

Now, we put the value of λ in equation (4), we get

= (2 + 15/2)x + (3 – 25/2)y + 1 – 25/2 = 0 and,

= (2 + 3/8)x + (3 – 5/8)y + 1 – 5/8 = 0

19x – 19y – 23 = 0 and 19x + 19y + 3 = 0

Hence, the equation of line is 19x – 19y – 23 = 0 and 19x + 19y + 3 = 0

### Question 5. Find the equation of the straight line drawn through the point of intersection of the lines x + y = 4 and 2x – 3y = 1 and perpendicular to the line cutting off intercepts 5, 6 on the axes.

**Solution:**

From the question we have,

The equations of lines which are :

x + y = 4 …..(1)

2x – 3y = 1 …..(2)

Now we find the equation of the straight line that pass through the

points of intersection of equations (1) and (2) is given,

x + y − 4 + λ(2x − 3y − 1) = 0 …(3)

(1 + 2λ)x + (1 − 3λ)y − 4 − λ = 0 ….(4)

y = – [(1 + 2λ) / (1 – 3λ)]x + [(4 + λ) / (1 – 3λ)]

It is given that the equation of the line with intercepts 5 and 6 on the axis is

x/5 + y/6 = 1 ……..(5)

and the slope is -6/5

The lines (4) and (5) are perpendicular so,

-6/5 × [(-1 + 2λ) / (1 – 3λ)] = -1

λ = 11/3

Now, put the values of λ in (1), we get

(1 + 2(11/3))x + (1 – 3(11/3))y − 4 – 11/3 = 0

(1 + 22/3)x + (1 – 11)y – 4 – 11/3 = 0

25x – 30y – 23 = 0

Hence, the equation of line is 25x – 30y – 23 = 0

### Question 6. Prove that the family of lines represented by x (1 + λ) + y (2 – λ) + 5 = 0, λ being arbitrary, pass through a fixed point. Also, find the fixed point.

**Solution:**

According to the question

The family of lines represented by x (1 + λ) + y (2 – λ) + 5 = 0

So, x + xλ + 2y – λy + 5 = 0

λ(x – y) + (x + 2y + 5) = 0

(x + 2y + 5) + λ(x – y) = 0

So, this is L

_{1}+ λL_{2 }= 0Hence, the line passing through the intersection of x – y = 0 and x + 2y = -5.

So, (-5/3, -5/3) which is the fixed point through which the lines pass for any value of λ.

### Question 7. Show that the straight lines given by (2 + k)x + (1 + k)y = 5 + 7k for different values of k pass through a fixed point. Also, find the point.

**Solution:**

From the question we have,

The equation of line is :

(2 + k) x + (1 + k) y = 5 + 7k

(2x + y – 5) + (x + y = 7) = 0

It is of the form l₁ + kL₂ = 0

So, it represents a line passing through:

2x + y – 5 = 0 ……(1)

x +y – 7 = 0 ……(2)

On Solving equation(1) and (2) we get,

The value of the point (-2, 9).

### Question 8. Find the equation of the straight line passing through the point of intersection of 2x + y – 1 = 0 and x + 3y – 2 = 0 and making with the coordinate axes a triangle of area 3/8 sq units.

**Solution:**

From the question we have,

The equations of lines which are :

2x + y – 1 = 0 …..(1)

x + 3y – 2 = 0 …..(2)

Now we find the equation of the straight line that pass through the

points of intersection of equations (1) and (2) is given,

(2x +y -1) + λ(x + 3y – 2) = 0 ……(3)

x (2 + λ) + y (1 + 3λ) – 1 – 2λ = 0

x/((1 + 2λ) / (2 + λ)) + 4/((1 + 2λ) / (1 + 3λ)) = 1

As we know that the area of the triangle OAB = 1/2 × OB × OA

8 / 3 =1/2 × (y intercept) × (x intercept)

8/3 = 1/2 × (1 + 2λ / 1 + 3λ) × (1 + 2λ / 2 + λ)

16/3 = (4λ

^{2 }+ 4λ) / (3λ + 3λ^{2}+ 7λ)60λ

^{2}+ 124λ + 35 = 0λ = -124 ± √(124)

^{2 }– 4×60 × 35 / 2×60λ = -124 ± √15376-8400 / 120

λ = 1(Approximately)

Now put the value of λ in (3) we get

3x + 4y -30 = 0,

Hence, the equation of line is 12x + y – 3 = 0

### Question 9. Find the equation of straight line passes through the point of intersection of the lines 3x – y = 5 and x + 3y = 1 and makes equal and positive intercepts on the axis.

**Solution:**

From the question we have,

The equations of lines which are :

3x – y = 5 …..(1)

x + 3y = 1 …..(2)

Now we find the equation of the straight line that pass through the

points of intersection of equations (1) and (2) is given,

(3x – y – 5) + λ(x + 3y – 1) = 0 …..(3)

x/(5 + λ/3 + λ) + y/(5 + λ/3λ – 1) = 1

It is given that the line which makes equal and positive intercepts with the line

So, 5 + λ /3 + λ = 5 + λ/3λ – 1

3λ – 1 = 3 + λ

2λ = 4

λ = 2

Now put the value of λ in (3) we get

3x – y – 5 + 2x + 6y – 2 = 0

Hence, the equation of line is 5x + 5y = 7

### Question 10. Find the equation of straight line passes through the point of intersection of the lines x – 3y + 1 = 0 and 2x – 5y – 9 and whose distance from the origin is√5.

**Solution:**

From the question we have,

The equations of lines which are :

x – 3y + 1 = 0 …..(1)

2x – 5y – 9 …..(2)

Now we find the equation of the straight line that pass through the

points of intersection of equations (1) and (2) is given,

x – 3y + 1 + λ(2x + 5y – 9) = 0 ….(3)

(1 + 2λ) x + (-3 + 5λ) y + 1 – 9λ = 0

It is given that the distance from the origin is√5

So,

D = |((1 + 2λ) 0 + (-3 + 5λ) 0 + 1 – 9λ) / √(1 + 2λ)

^{2}+ (5λ – 3)^{2}|√5 = | 1 – 9λ / √1 + 4λ

^{2}+ 4λ + 25λ^{2}+ 9 – 30λ |5(10 + 29λ

^{2 }– 26λ) = (1 – 9λ)^{2}50 + 145λ

^{2 }– 130λ = 1 + 81λ^{2 }– 18λ^{2}64λ

^{2 }– 112λ + 49 = 0(8λ – 7)

^{2}= 0λ =7/8

Now put the value of λ in (3) we get

= x – 3y + 1+ (7/8) (2x + 5y – 9) = 0

8x – 24y + 8 + 14x + 35y – 63 = 0

22x + 11y – 55 = 0

Hence, the equation of line is 2x + y – 5 = 0

### Question 11. Find the equations of the lines through the point of intersection of the lines x – y + 1 = 0 and 2x – 3y + 5 = 0 whose distance from the point (3, 2) is 7/5.

**Solution:**

From the question we have,

The equations of lines which are :

x – y + 1 = 0 …..(1)

2x – 3y + 5 = 0 …..(2)

On solving these two equations of lines we get,

intersection point (2, 3).

Now, let equation of a line passing through (2, 3) be

y – mx + c

3 = 2m + c

c = 3 – 2m

So, the equation of the line is y – mx + 3 – 2m ……(3)

It is given that the distance from the point (3, 2) is 7/5.

|3m – 2 + 3 – 2m / √m

^{2 }+ 1 | = 7/5| m + 1 / √m

^{2}+1 | = 7/5(m + 1)

^{2 }/ m^{2}+ 1 = 49 / 2525(m

^{2}+ 2m + 1) = 49m^{2}+ 4925m

^{2 }+ 50m + 25 = 49m^{2}+ 4924m

^{2}– 50m + 24 = 012m

^{2}– 25m + 12 = 0m = 4/3, m = 3 /4

Now put the value of m in eq(3), we get

y = (4/3)x + 3 – 8/3

3y = 4x + 1

4x – 3y + 1 = 0

y = (3/ 4)x +3 – 6/4

4y – 3y + 1 = 0

Hence, the equations of lines are 4x – 3y + 1 = 0 and 4y – 3y + 1 = 0

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