# Class 11 RD Sharma Solutions – Chapter 23 The Straight Lines – Exercise 23.4

Last Updated : 18 Mar, 2021

### Question 1. Find the equation of straight line passing through the point (6, 2) and having slope -3.

Solution:

We know (y – y1) = m(x – x1)

Here, m = Slope of line = -3

x1 = 6, y1 = 2

So, the equation of line is

â‡’ y – 2 = (-3)(x – 6)

â‡’ y – 2 = -3x + 18

â‡’ 3x + y – 20 = 0

### Question 2. Find the equation of straight line passing through (-2, 3) and inclined at an angle of 45Â° with the x-axis.

Solution:

We know y – y1 = m(x – x1)

Here, m = Slope of line = tan 45Â° = 1

x1 = -2, y1 = 3

So, the equation of line is

â‡’ y – 3 = 1(x – (-2))

â‡’ y – 3 = x + 2

â‡’ x – y + 5 = 0

### Question 3. Find the equation of straight line which divides the join of points (2, 3) and (-5, 8) in the ratio 3:4 and is also perpendicular to it.

Solution:

Let the point at which the line divides the join of points (2, 3) and (-5, 8) in the ratio 3:4 be P(x1, y1)

P(x1, y1) = \frac{(4 × 2 - 5 × 3)}{(3 + 4)}, \frac{(4 × 3 + 3 × 8)}{(3+4)}

= (-1, 36/7)

Slope of the given points=(8-3)/(-5-2) =-5/7

Since, the required line is perpendicular to the line joining the given points.

Therefore, the slope of required line ‘m’ = 7/5

Here, x1 = -1, y1 = 36/7 & m = 7/5

So, the equation of line is

â‡’ y – 36/7 = 7/5(x – (-1))

â‡’ y-36/7 = 7/5(x + 1)

â‡’ 35y – 180 = 49x + 49

â‡’ 49x + 35y + 229 = 0 is the required equation of straight line.

### Question 4. Prove that the perpendicular drawn from the point (4, 1) on the join of (2, -1) and (6, 5) divides it in ratio 5:8.

Solution:

Let PO be the perpendicular drawn from P(4,1) on the line joining A(2, -1) and B(6, 5)

Let slope of PO be ‘m’

According to question

m Ã— Slope of AB = -1

â‡’ m Ã— (5 + 1)/(6 – 2) = -1

â‡’ m Ã— 6/4 = -1

â‡’ m = -4/6 = -2/3

Thus, the equation of line PO

x1 = 4, y1 = 1 & m = -2/3

(y – 1) = -2/3(x – 4)

â‡’ 3y – 3 = -2x + 8

â‡’ 2x + 3y – 11 = 0 ——–(1)

Let O divide the line AB in the ratio of K:1

Then the coordinates of O are \frac{(6k + 2)}{(k + 1)}, \frac{(5k - 1)}{(k + 1)}

Since point O lies in the line AB

Therefore, it satisfies the equation (1)

â‡’ 12k + 4 + 15k – 3 – 11(k + 1) = 0

â‡’ 27k – 11k + 1 – 11 = 0

â‡’ 16k = 10

â‡’ k = 5/8

Hence Proved

### Question 5. Find the equation to the altitudes of the triangle whose angular points are A(2, -2), B(1, 1), and C(-1, 0).

Solution:

Let the altitudes be AE, BF and CD

Slope of AE Ã— Slope of BC = -1                       [Since both lines are perpendicular to each other]

Slope of AE = (-1) / Slope of BC

Slope of AE =

= -2

Equation of the altitude AE

y – (-2) = (-2)(x – 2)

â‡’ y + 2 = -2x + 4

â‡’ 2x + y – 2 = 0

Similarly,

Slope of BF Ã— Slope of AC = -1

Slope of BF =

= 3/2

Equation of the altitude BF

y – 1 = (3/2)(x – 1)

â‡’ 2(y – 1) = 3x – 3

â‡’ -3x + 2y – 2 + 3 = 0

â‡’ 3x – 2y – 1 = 0

Slope of CD Ã— Slope of AB = -1

Slope of CD=

= 1/3

Equation of the altitude CD

y – 0 = (1/3) (x – (-1))

â‡’ 3y = x + 1

â‡’ x – 3y + 1 = 0

### Question 6. Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).

Solution:

Let the points of the line segment be A(3, 4) and B(-1, 2)

Let the right bisector meet at point ‘P’ on the line segment

Coordinates of point P = \frac{(3-1)}{2}, \frac{(4+2)}{2}          [Using the mid point formula]

= (1, 3)

Slope of AB = [(2 – 4) / (-1 – 3)]

=-2/-4

= 1/2

Slope of right bisector =

= -2

Equation of the right bisector

â‡’ y – 3 = -2(x – 1)

â‡’ y – 3 = -2x + 2

â‡’ 2x + y – 5 = 0 is the required equation of the right bisector

### Question 7. Find the equation of the straight line passing through the point (3, -2) and making an angle of 60Â° with the positive direction y-axis.

Solution:

Since the line is making an angle of 60Â° with the positive direction of y-axis i

t makes an angle of 30Â° with the positive direction of x-axis as shown in the diagram.

Slope of line ‘m’= tan 30Â° = 1/âˆš3

Equation of straight line passing through (3, -2)

y – (-2) = 1/âˆš3(x – 3)

â‡’ âˆš3(y + 2) = x – 3

â‡’ âˆš3y + 2âˆš3 = x – 3

â‡’ x – âˆš3y – 3 – 2âˆš3 = 0 is the required equation of straight line.

### Question 8. Find the equation of the straight line which passes through the point (1, 2) and makes an angle with the positive direction of x-axis whose sine is 3/5.

Solution:

Given, sin Î¸ = 3/5

tan Î¸ = 3/âˆš(25 – 9) = 3/4

Slope of the line m = 3/4

Equation of straight line passing through (1, 2)

y – 2 = 3/4(x – 1)

â‡’ 4(y – 2) = 3x – 3

â‡’ 4y – 8 = 3x – 3

â‡’ 3x – 4y + 5 = 0 is the required equation of straight line.

### Question 9. Find the equation of the line passing through the point (-3, 5) and perpendicular to the line joining (2, 5) and (-3, 6).

Solution:

Slope of the required equation ‘m’ = (-1) / [Slope of line joining (2, 5) and (-3, 6)]                 [Perpendicular to each other]

m =

m =

m = 5

Equation of straight line passing through (-3, 5)

y – 5 = 5(x – (-3))

â‡’ y – 5 = 5x + 15

â‡’ 5x – y + 20 = 0 is the required equation of straight line.

### Question 10. Find the equation of the right bisector of the line segment joining the points A(1, 0) and B(2, 3).

Solution:

Let the right bisector meet at point ‘P’ on the line segment

Coordinates of point P = [ ]                   [Using the mid point formula]

= (3/2, 3/2)

Slope of AB = [(3 – 0)/(2 – 1)]

= 3/1 = 3

Slope of right bisector = -1/3

Equation of the right bisector

â‡’ y – 3/2 = -1/3(x – 3/2)

â‡’ 3(y – 3/2) = -x + 3/2

â‡’ x + 3y – 9/2 – 3/2 = 0

â‡’ x + 3y – 6 = 0

â‡’ x + 3y – 6 = 0 is the required equation of  the right bisector.

### Question 11. Find the lines through the point (0, 2) making angles Ï€/3 and 2Ï€/3 with the x-axis. Also, find the lines parallel to them cutting the y-axis at a distance of 2 units below the origin.

Solution:

Given the line through the point (0, 2) making angles Ï€/3 and 2Ï€/3 with the x-axis

.Slope m1= tan Ï€/3 = âˆš3

Slope m2 = tan 2Ï€/3 = -âˆš3

Equation of the required lines

â‡’ y – 2 = âˆš3(x – 0) and y – 2 = -âˆš3(x – 0)

â‡’ y – âˆš3x – 2 = 0 and y + âˆš3x – 2 = 0

Now, the equation of the line parallel to the line having slope m1 and y intercept c= -2

y = m1x + c

â‡’ y = âˆš3x – 2

Similarly, the equation of the line parallel to the line having slope m2 and y intercept c = -2

y= m2x + c

â‡’ y = -âˆš3x – 2

### Question 12.  Find the equation of the straight lines which cut off an intercept 5 from the y-axis and are equally inclined to the axes.

Solution:

Given that the straight lines cut off an intercept 5 from the y-axis and are equally inclined to the axes.

Slope of the two lines are m1= tan 45Â° =1 and m2 = tan 135Â° = -1

Equation of the required straight lines are

y = m1x + c or y = m2 + c

â‡’ y = x +5 or y = -x + 5

â‡’ y = x +5 or y + x = 5

### Question 13.  Find the equation of straight lines which intercepts a length 2 on the positive direction of the x-axis and is inclined at an angle of 135Â° with the positive direction of y-axis.

Solution:

The required line which is inclined at an angle of 135Â° with the positive direction of y-axis makes an angle of 45Â° with the positive x-axis.

Slope of the required line m = tan 45Â° = -1

Equation of the required straight line with x-intercept c = 2 and m = -1

x = my + c

â‡’ x = 1y + 2

â‡’ x – y  – 2 = 0

### Question 14. Find the equation of line passing through (0, 0) with slope m.

Solution:

Equation of line passing through (0, 0) with slope m is

y – 0 = m(x – 0)

â‡’ y = mx

### Question 15. Find the equation of line passing through (2, 2âˆš3) and inclined with x-axis at an angle of 75Â°.

Solution:

Slope of the line m = tan 75Â°

= tan (45Â° + 30Â°)

= (tan 45Â° + tan30Â°) /(1 – tan 45Â° tan30Â°)

= (1 + 1/âˆš3)/(1 – 1/âˆš3)

m = (âˆš3 + 1)/(âˆš3 – 1) = 2 + âˆš3

Equation of the required line passing through (2, 2âˆš3) with slope of 2 + âˆš3

y – 2âˆš3 = (2 + âˆš3)(x – 2)

â‡’ y – 2âˆš3  = (2 + âˆš3)x – 4 – 2âˆš3

â‡’ (2 + âˆš3)x – y – 4 = 0

### Question 16. Find the equation of the line passing through(1, 2) and making angle of 30Â° with the y-axis.

Solution:

Let us considered an equation of line passing through points(x1, y1) which making an angle Î¸ with x-axis.

(y – y1) = tanÎ¸(x – x1)         -(1)

Given: Point = (1, 2), and angle = 30Â°(with y-axis)

So, angle with x-axis = 90Â° – 30Â° = 60Â°

Now put all these values in eq(1), we get

(y – 2) = tan60Â°(x – 1)

(y – 2) = âˆš3(x – 1)

y – 2 = âˆš3x – âˆš3

âˆš3x – âˆš3 – y + 2 = 0

âˆš3x – y – âˆš3  + 2 = 0

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