RD Sharma Solutions for Class 11 Chapter 23 - The Straight Lines

“A straight line is a curve such that every point on the line segment joining any two points lies on it is termed as a straight line”. In this chapter, we shall discuss concepts related to Straight lines, with examples for better understanding. RD Sharma Class 11 Maths Solutions contains all the solutions to the Maths problems provided in the textbook. The subject experts have framed and solved the questions accurately from every section. RD Sharma Solutions for Class 11 is a detailed and step-by-step guide to all the queries of the students. The exercises present in the chapter should be dealt with utmost sincerity, if one aims to score well in the examinations. Students can refer and download this chapter pdf from the below-mentioned links and start practising offline.

Chapter 23 – The Straight Lines contains nineteen exercises and the RD Sharma Solutions present in this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

  • Definition of a straight line.
  • Slope (Gradient) of a line.
  • Angle between two lines.
  • Intercepts of a line on the axes.
  • Equations of lines parallel to the coordinate axes.
  • Different forms of the equation of a straight line.
    • Slope intercept form of a line.
    • Point-slope form of a line.
    • Two-point form of a line.
    • The intercept form of a line.
    • Normal form of perpendicular form of a line.
    • Distance form of a line.
  • Transformation of general equations in different standard forms.
  • Point of intersection of two lines.
  • Condition of concurrency of three lines.
  • Lines parallel and perpendicular to a given line.
  • Angle between two straight lines when their equations are given.
  • Position of two points relative to a line.
  • Distance of a point from a line.
  • Distance between parallel lines.
  • Area of parallelogram.
  • Equations of lines passing through a given point and making a given angle with a line.
  • Family of lines through the intersection of two given lines.

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Access answers to RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines

EXERCISE 23.1 PAGE NO: 23.12

1. Find the slopes of the lines which make the following angles with the positive direction of x – axis:

(i) – π/4

(ii) 2π/3

Solution:

(i) – π/4

Let the slope of the line be ‘m’

Where, m = tan θ

So, the slope of Line is m = tan (– π/4) 

= – 1

∴ The slope of the line is – 1.

(ii) 2π/3

Let the slope of the line be ‘m’

Where, m = tan θ

So, the slope of Line is m = tan (2π/3)

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 1

∴ The slope of the line is –3

2. Find the slopes of a line passing through the following points :
(i) (–3, 2) and (1, 4)

(ii) (at21, 2at1) and (at22, 2at2)

Solution:

(i) (–3, 2) and (1, 4)

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 2

∴ The slope of the line is ½.

(ii) (at21, 2at1) and (at22, 2at2)

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 3

3. State whether the two lines in each of the following are parallel, perpendicular or neither:
(i) Through (5, 6) and (2, 3); through (9, –2) and (6, –5)

(ii) Through (9, 5) and (– 1, 1); through (3, –5) and (8, –3)

Solution:

(i) Through (5, 6) and (2, 3); through (9, –2) and (6, –5)

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 4

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 5

So, m2 = 1

Here, m1 = m2 = 1

∴ The lines are parallel to each other.

(ii) Through (9, 5) and (– 1, 1); through (3, –5) and (8, –3)

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 6

4. Find the slopes of a line
(i) which bisects the first quadrant angle
(ii) which makes an angle of 300 with the positive direction of y – axis measured anticlockwise.

Solution:

(i) Which bisects the first quadrant angle?

Given: Line bisects the first quadrant

We know that, if the line bisects in the first quadrant, then the angle must be between line and the positive direction of x – axis.

Since, angle = 90/2 = 45o 

By using the formula,

The slope of the line, m = tan θ

The slope of the line for a given angle is m = tan 45o

So, m = 1

∴ The slope of the line is 1.


(ii) Which makes an angle of 300 with the positive direction of y – axis measured anticlockwise?

Given: The line makes an angle of 30o with the positive direction of y – axis.

We know that, angle between line and positive side of axis => 90o + 30o = 120o

By using the formula,

The slope of the line, m = tan θ

The slope of the line for a given angle is m = tan 120o

So, m = – √3

∴ The slope of the line is – √3.

5. Using the method of slopes show that the following points are collinear:
(i) A (4, 8), B (5, 12), C (9, 28)

(ii) A(16, – 18), B(3, – 6), C(– 10, 6)

Solution:

(i) A (4, 8), B (5, 12), C (9, 28)

By using the formula,

The slope of the line = [y2 – y1] / [x2 – x1]

So,

The slope of line AB = [12 – 8] / [5 – 4] 

= 4 / 1

The slope of line BC = [28 – 12] / [9 – 5] 

= 16 / 4

= 4

The slope of line CA = [8 – 28] / [4 – 9] 

= -20 / -5

= 4

Here, AB = BC = CA

∴ The Given points are collinear.

(ii) A(16, – 18), B(3, – 6), C(– 10, 6)

By using the formula,

The slope of the line = [y2 – y1] / [x2 – x1]

So,

The slope of line AB = [-6 – (-18)] / [3 – 16] 

= 12 / -13

The slope of line BC = [6 – (-6)] / [-10 – 3] 

= 12 / -13

The slope of line CA = [6 – (-18)] / [-10 – 16] 

= 12 / -13

= 4

Here, AB = BC = CA

∴ The Given points are collinear.

EXERCISE 23.2 PAGE NO: 23.17

1. Find the equation of the parallel to x–axis and passing through (3, –5).

Solution:

Given: A line which is parallel to x–axis and passing through (3, –5)

By using the formula,

The equation of line: [y – y1 = m(x – x1)]

We know that the parallel lines have equal slopes

And, the slope of x–axis is always 0

Then

The slope of line, m = 0

Coordinates of line are (x1, y1) = (3, –5)

The equation of line = y – y1 = m(x – x1)

Now, substitute the values, we get

y – (– 5) = 0(x – 3)

y + 5 = 0

∴ The equation of line is y + 5 = 0

2. Find the equation of the line perpendicular to x–axis and having intercept – 2 on x–axis.

Solution:

Given: A line which is perpendicular to x–axis and having intercept –2

By using the formula,

The equation of line: [y – y1 = m(x – x1)]

We know that, the line is perpendicular to the x–axis, then x is 0 and y is –1.

The slope of line is, m = y/x 

= -1/0

It is given that x–intercept is –2, so, y is 0.

Coordinates of line are (x1, y1) = (–2, 0)

The equation of line = y – y1 = m(x – x1)

Now, substitute the values, we get

y – 0 = (-1/0) (x – (– 2))

x + 2 = 0

∴ The equation of line is x + 2 = 0

3. Find the equation of the line parallel to x–axis and having intercept – 2 on y – axis.

Solution:

Given: A line which is parallel to x–axis and having intercept –2 on y – axis

By using the formula,

The equation of line: [y – y1 = m(x – x1)]

The parallel lines have equal slopes,

And, the slope of x–axis is always 0

Then

The slope of line, m = 0

It is given that intercept is –2, on y – axis then

Coordinates of line are (x1, y1) = (0, – 2)

The equation of line is y – y1 = m(x – x1)

Now, substitute the values, we get

y – (– 2) = 0 (x – 0)

y + 2 = 0

∴ The equation of line is y + 2 = 0

4. Draw the lines x = –3, x = 2, y = –2, y = 3 and write the coordinates of the vertices of the square so formed.

Solution:

Given: x = –3, x = 2, y = –2 and y = 3

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 7

∴ The Coordinates of the square are: A(2, 3), B(2, –2), C(–3, 3), and D(–3, –2).

5. Find the equations of the straight lines which pass through (4, 3) and are respectively parallel and perpendicular to the x–axis.

Solution:

Given: A line which is perpendicular and parallel to x–axis respectively and passing through (4, 3)

By using the formula,

The equation of line: [y – y1 = m(x – x1)]

Let us consider,

Case 1: When Line is parallel to x–axis

The parallel lines have equal slopes,

And, the slope of x–axis is always 0, then

The slope of line, m = 0

Coordinates of line are (x1, y1) = (4, 3)

The equation of line is y – y1 = m(x – x1)

Now substitute the values, we get

y – (3) = 0(x – 4)

y – 3 = 0

Case 2: When line is perpendicular to x–axis

The line is perpendicular to the x–axis, then x is 0 and y is – 1.

The slope of the line is, m = y/x

= -1/0

Coordinates of line are (x1, y1) = (4, 3)

The equation of line = y – y1 = m(x – x1)

Now substitute the values, we get

y – 3 = (-1/0) (x – 4)

x = 4

∴ The equation of line when it is parallel to x – axis is y = 3 and it is perpendicular is x = 4.

EXERCISE 23.3 PAGE NO: 23.21

1. Find the equation of a line making an angle of 150° with the x–axis and cutting off an intercept 2 from y–axis.

Solution:

Given: A line which makes an angle of 150o with the x–axis and cutting off an intercept at 2

By using the formula,

The equation of a line is y = mx + c

We know that angle, θ = 150o

The slope of the line, m = tan θ

Where, m = tan 150o

= -1/3

Coordinate of y–intercept is (0, 2)

The required equation of the line is y = mx + c

Now substitute the values, we get

y = -x/3 + 2

3y – 23 + x = 0

x + 3y = 23

∴ The equation of line is x + 3y = 23

2. Find the equation of a straight line:
(i) with slope 2 and y – intercept 3;
(ii) with slope – 1/ 3 and y – intercept – 4.
(iii) with slope – 2 and intersecting the x–axis at a distance of 3 units to the left of origin.

Solution:

(i) With slope 2 and y – intercept 3

The slope is 2 and the coordinates are (0, 3)

Now, the required equation of line is

y = mx + c

Substitute the values, we get

y = 2x + 3

(ii) With slope – 1/ 3 and y – intercept – 4

The slope is – 1/3 and the coordinates are (0, – 4)

Now, the required equation of line is

y = mx + c

Substitute the values, we get

y = -1/3x – 4

3y + x = – 12


(iii) With slope – 2 and intersecting the x–axis at a distance of 3 units to the left of origin

The slope is – 2 and the coordinates are (– 3, 0)

Now, the required equation of line is y – y1 = m (x – x1)

Substitute the values, we get

y – 0 = – 2(x + 3)

y = – 2x – 6

2x + y + 6 = 0

3. Find the equations of the bisectors of the angles between the coordinate axes.

Solution:

There are two bisectors of the coordinate axes.

Their inclinations with the positive x-axis are 45o and 135o

The slope of the bisector is m = tan 45o or m = tan 135o

i.e., m = 1 or m = -1, c = 0

By using the formula, y = mx + c

Now, substitute the values of m and c, we get

y = x + 0

x – y = 0 or y = -x + 0

x + y = 0

∴ The equation of the bisector is x ± y = 0

4. Find the equation of a line which makes an angle of tan – 1 (3) with the x–axis and cuts off an intercept of 4 units on the negative direction of y–axis.

Solution:

Given:

The equation which makes an angle of tan – 1(3) with the x–axis and cuts off an intercept of 4 units on the negative direction of y–axis

By using the formula,

The equation of the line is y = mx + c

Here, angle θ = tan – 1(3)

So, tan θ = 3

The slope of the line is, m = 3

And, Intercept in the negative direction of y–axis is (0, -4)

The required equation of the line is y = mx + c

Now, substitute the values, we get

y = 3x – 4

∴ The equation of the line is y = 3x – 4.

5. Find the equation of a line that has y – intercept – 4 and is parallel to the line joining (2, –5) and (1, 2).

Solution:

Given:

A line segment joining (2, – 5) and (1, 2) if it cuts off an intercept – 4 from y–axis

By using the formula,

The equation of line is y = mx + C

It is given that, c = – 4

Slope of line joining (x1 – x2) and (y1 – y2),

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 8

So, Slope of line joining (2, – 5) and (1, 2),

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 9

m = – 7

The equation of line is y = mx + c

Now, substitute the values, we get

y = –7x – 4

y + 7x + 4 = 0

∴ The equation of line is y + 7x + 4 = 0.

EXERCISE 23.4 PAGE NO: 23.29

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

Solution:

Given, A straight line passing through the point (6, 2) and the slope is – 3

By using the formula,

The equation of line is [y – y1 = m(x – x1)]

Here, the line is passing through (6, 2)

It is given that, the slope of line, m = –3

Coordinates of line are (x1, y1) = (6,2)

The equation of line = y – y1 = m(x – x1)

Now, substitute the values, we get

y – 2 = – 3(x – 6)

y – 2 = – 3x + 18

y + 3x – 20 = 0

∴ The equation of line is 3x + y – 20 = 0

2. Find the equation of the straight line passing through (–2, 3) and indicated at an angle of 45° with the x – axis.

Solution:

Given:

A line which is passing through (–2, 3), the angle is 45o.

By using the formula,

The equation of line is [y – y1 = m(x – x1)]

Here, angle, θ = 45o

The slope of the line, m = tan θ

m = tan 45o

= 1

The line passing through (x1, y1) = (–2, 3)

The required equation of line is y – y1 = m(x – x1)

Now, substitute the values, we get

y – 3 = 1(x – (– 2))

y – 3 = x + 2

x – y + 5 = 0

∴The equation of line is x – y + 5 = 0

3. Find the equation of the line passing through (0, 0) with slope m

Solution:

Given:

A straight line passing through the point (0, 0) and slope is m.

By using the formula,

The equation of line is [y – y1 = m(x – x1)]

It is given that, the line is passing through (0, 0) and the slope of line, m = m

Coordinates of line are (x1, y1) = (0, 0)

The equation of line = y – y1 = m(x – x1)

Now, substitute the values, we get

y – 0 = m(x – 0)

y = mx

∴ The equation of line is y = mx.

4. Find the equation of the line passing through (2, 2√3) and inclined with x – axis at an angle of 75o.

Solution:

Given:

A line which is passing through (2, 2√3), the angle is 75o.

By using the formula,

The equation of line is [y – y1 = m(x – x1)]

Here, angle, θ = 75o

The slope of the line, m = tan θ

m = tan 75o

= 3.73 = 2 + √3

The line passing through (x1, y1) = (2, 2√3)

The required equation of the line is y – y1 = m(x – x1)

Now, substitute the values, we get

y – 2√3 = 2 + √3 (x – 2)

y – 2√3 = (2 + √3)x – 7.46

(2 + √3)x – y – 4 = 0

∴ The equation of the line is (2 + √3)x – y – 4 = 0

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

Solution:

A line which is passing through (1, 2)

To Find: The equation of a straight line.

By using the formula,

The equation of line is [y – y1 = m(x – x1)]

Here, sin θ = 3/5

We know, sin θ = perpendicular/hypotenuse

= 3/5

So, according to Pythagoras theorem,

(Hypotenuse)2 = (Base)2 + (Perpendicular)2

(5)2 = (Base)2 + (3)2

(Base) = √(25 – 9)

(Base)2 = √16 

Base = 4

Hence, tan θ = perpendicular/base

= 3/4

The slope of the line, m = tan θ

= 3/4

The line passing through (x1,y1) = (1,2)

The required equation of line is y – y1 = m(x – x1)

Now, substitute the values, we get

y – 2 = (¾) (x – 1)

4y – 8 = 3x – 3

3x – 4y + 5 = 0

∴ The equation of line is 3x – 4y + 5 = 0

EXERCISE 23.5 PAGE NO: 23.35

1. Find the equation of the straight lines passing through the following pair of points:
(i) (0, 0) and (2, -2)

(ii) (a, b) and (a + c sin α, b + c cos α)

Solution:

(i) (0, 0) and (2, -2)

Given:

(x1, y1) = (0, 0), (x2, y2) = (2, -2)

The equation of the line passing through the two points (0, 0) and (2, −2) is

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 10

y = – x

∴ The equation of line is y = -x

(ii) (a, b) and (a + c sin α, b + c cos α)

Given:

(x1, y1) = (a, b), (x2, y2) = (a + c sin α, b + c cos α)

So, the equation of the line passing through the two points (0, 0) and (2, −2) is

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 11

y – b = cot α (x – a)

∴ The equation of line is y – b = cot α (x – a)

2. Find the equations to the sides of the triangles the coordinates of whose angular points are respectively:
(i) (1, 4), (2, -3) and (-1, -2)

(ii) (0, 1), (2, 0) and (-1, -2)

Solution:

(i) (1, 4), (2, -3) and (-1, -2)

Given:

Points A (1, 4), B (2, -3) and C (-1, -2).

Let us assume,

m1, m2, and m3 be the slope of the sides AB, BC and CA, respectively.

So,

The equation of the line passing through the two points (x1, y1) and (x2, y2).

Then,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 12

m1 = -7, m2 = -1/3 and m3 = 3

So, the equation of the sides AB, BC and CA are

By using the formula,

y – y1= m (x – x1)

=> y – 4 = -7 (x – 1)

y – 4 = -7x + 7

y + 7x = 11,

=> y + 3 = (-1/3) (x – 2)

3y + 9 = -x + 2

3y + x = – 7

x + 3y + 7 = 0 and

=> y + 2 = 3(x+1)

y + 2 = 3x + 3

y – 3x = 1

So, we get

y + 7x =11, x+ 3y + 7 =0 and y – 3x = 1

∴ The equation of sides are y + 7x =11, x+ 3y + 7 =0 and y – 3x = 1

(ii) (0, 1), (2, 0) and (-1, -2)

Given:

Points A (0, 1), B (2, 0) and C (-1, -2).

Let us assume,

m1, m2, and m3 be the slope of the sides AB, BC and CA, respectively.

So,

The equation of the line passing through the two points (x1, y1) and (x2, y2).

Then,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 13

m1 = -1/2, m2 = -2/3 and m3= 3

So, the equation of the sides AB, BC and CA are

By using the formula,

y – y1= m (x – x1)

=> y – 1 = (-1/2) (x – 0)

2y – 2 = -x

x + 2y = 2

=> y – 0 = (-2/3) (x – 2)

3y = -2x + 4

2x – 3y = 4

=> y + 2 = 3(x+1)

y + 2 = 3x + 3

y – 3x = 1

So, we get

x + 2y = 2, 2x – 3y =4 and y – 3x = 1

∴ The equation of sides are x + 2y = 2, 2x – 3y =4 and y – 3x = 1

3. Find the equations of the medians of a triangle, the coordinates of whose vertices are (-1, 6), (-3,-9) and (5, -8).

Solution:

Given:

A (−1, 6), B (−3, −9) and C (5, −8) be the coordinates of the given triangle.

Let us assume: D, E, and F be midpoints of BC, CA and AB, respectively. So, the coordinates of D, E and F are

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 14

Median AD passes through A (-1, 6) and D (1, -17/2)

So, by using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 15

4y – 24 = -29x – 29

29x + 4y + 5 = 0

Similarly, Median BE passes through B (-3,-9) and E (2,-1)

So, by using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 16

5y + 45 = 8x + 24

8x – 5y – 21=0


Similarly, Median CF passes through C (5,-8) and F(-2,-3/2)

So, by using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 17

 -14y – 112 = 13x – 65

13x + 14y + 47 = 0

∴ The equation of lines are: 29x + 4y + 5 = 0, 8x – 5y – 21=0 and 13x + 14y + 47 = 0

4. Find the equations to the diagonals of the rectangle the equations of whose sides are x = a, x = a’, y = b and y = b’.

Solution:

Given:

The rectangle formed by the lines x = a, x = a’, y = b and y = b’

It is clear that, the vertices of the rectangle are A (a, b), B (a’, b), C (a’, b’) and D (a, b’) .

The diagonal passing through A (a, b) and C (a’, b’) is

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 18

(a’ – a)y – b(a’ – a) = (b’ – b)x – a(b’ – b)

(a’ – a) – (b’ – b)x = ba’ – ab’


Similarly, the diagonal passing through B (a’, b) and D (a, b’) is

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 19

(a’ – a)y – b(a – a’) = (b’ – b)x – a’ (b’ – b)

(a’ – a) + (b’ – b)x = a’b’ – ab

∴ The equation of diagonals are y(a’ – a) – x(b’ – b) = a’b – ab’ and

y(a’ – a) + x(b’ – b) = a’b’ – ab

5. Find the equation of the side BC of the triangle ABC whose vertices are A (-1, -2), B (0, 1) and C (2, 0) respectively. Also, find the equation of the median through A (-1, -2).

Solution:

Given:

The vertices of triangle ABC are A (-1, -2), B(0, 1) and C(2, 0).

Let us find the equation of median through A.

So, the equation of BC is

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 20

x + 2y – 2 = 0


Let D be the midpoint of median AD,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 21

4y + 8 = 5x + 5

5x – 4y – 3 = 0

∴ The equation of line BC is x + 2y – 2 = 0

The equation of median is 5x – 4y – 3 = 0

EXERCISE 23.6 PAGE NO: 23.46

1. Find the equation to the straight line
(i) cutting off intercepts 3 and 2 from the axes.

(ii) cutting off intercepts -5 and 6 from the axes.

Solution:

(i) Cutting off intercepts 3 and 2 from the axes.

Given:

a = 3, b = 2

Let us find the equation of line cutoff intercepts from the axes.

By using the formula,

The equation of the line is x/a + y/b = 1

x/3 + y/2 = 1

By taking LCM,

2x + 3y = 6

∴ The equation of line cut off intercepts 3 and 2 from the axes is 2x + 3y = 6

(ii) Cutting off intercepts -5 and 6 from the axes.

Given:

a = -5, b = 6

Let us find the equation of line cutoff intercepts from the axes.

By using the formula,

The equation of the line is x/a + y/b = 1

x/-5 + y/6 = 1

By taking LCM,

6x – 5y = -30

∴ The equation of line cut off intercepts 3 and 2 from the axes is 6x – 5y = -30

2. Find the equation of the straight line which passes through (1, -2) and cuts off equal intercepts on the axes.

Solution:

Given:

A line passing through (1, -2)

Let us assume, the equation of the line cutting equal intercepts at coordinates of length ‘a’ is

By using the formula,

The equation of the line is x/a + y/b = 1

x/a + y/a = 1

x + y = a

The line x + y = a passes through (1, -2)

Hence, the point satisfies the equation.

1 -2 = a

a = -1

∴ The equation of the line is x+ y = -1

3. Find the equation to the straight line which passes through the point (5, 6) and has intercepts on the axes
(i) Equal in magnitude and both positive

(ii) Equal in magnitude but opposite in sign

Solution:

(i) Equal in magnitude and both positive

Given:

a = b

Let us find the equation of line cutoff intercepts from the axes.

By using the formula,

The equation of the line is x/a + y/b = 1

x/a + y/a = 1

x + y = a

The line passes through the point (5, 6)

Hence, the equation satisfies the points.

5 + 6 = a

a = 11

∴ The equation of the line is x + y = 11

(ii) Equal in magnitude but opposite in sign

Given:

b = -a

Let us find the equation of line cutoff intercepts from the axes.

By using the formula,

The equation of the line is x/a + y/b = 1

x/a + y/-a = 1

x – y = a

The line passes through the point (5, 6)

Hence, the equation satisfies the points.

5 – 6 = a

a = -1

∴ The equation of the line is x – y = -1

4. For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x – 3y + 6 = 0 on the axes.

Solution:

Given:

Intercepts cut off on the coordinate axes by the line ax + by +8 = 0 …… (i)

And are equal in length but opposite in sign to those cut off by the line

2x – 3y +6 = 0 ……(ii)

We know that, the slope of two lines is equal

The slope of the line (i) is –a/b 

The slope of the line (ii) is 2/3

So let us equate,

-a/b = 2/3

a = -2b/3

The length of the perpendicular from the origin to the line (i) is

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 22

The length of the perpendicular from the origin to the line (ii) is

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 23

It is given that, d1 = d2

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 24

b = 4

So, a = -2b/3

= -8/3

∴ The value of a is -8/3 and b is 4.

5. Find the equation to the straight line which cuts off equal positive intercepts on the axes and their product is 25.

Solution:

Given:

a = b and ab = 25

Let us find the equation of the line which cutoff intercepts on the axes.

∴ a2 = 25

a = 5 [considering only positive value of intercepts]

By using the formula,

The equation of the line with intercepts a and b is x/a + y/b = 1 

x/5 + y/5 = 1

By taking LCM

x + y = 5

∴ The equation of line is x + y = 5

EXERCISE 23.7 PAGE NO: 23.53

1. Find the equation of a line for which

(i) p = 5, α = 60°

(ii) p = 4, α = 150°

Solution:

(i) p = 5, α = 60°

Given:

p = 5, α = 60°

The equation of the line in normal form is given by

Using the formula,

x cos α + y sin α = p

Now, substitute the values, we get

x cos 60° + y sin 60° = 5

x/2 + 3y/2 = 5

x + √3y = 10

∴ The equation of line in normal form is x + √3y = 10.

(ii) p = 4, α = 150°

Given:

p = 4, α = 150°

The equation of the line in normal form is given by

Using the formula,

x cos α + y sin α = p

Now, substitute the values, we get

x cos 150° + y sin 150° = 4

cos (180° – θ) = – cos θ , sin (180° – θ) = sin θ

x cos(180° – 30°) + y sin(180° – 30°) = 4

– x cos 30° + y sin 30° = 4

3x/2 + y/2 = 4

-√3x + y = 8

∴ The equation of line in normal form is -√3x + y = 8.

2. Find the equation of the line on which the length of the perpendicular segment from the origin to the line is 4 and the inclination of the perpendicular segment with the positive direction of x–axis is 30°.

Solution:

Given:

p = 4, α = 30°

The equation of the line in normal form is given by

Using the formula,

x cos α + y sin α = p

Now, substitute the values, we get

x cos 30° + y sin 30° = 4

x3/2 + y1/2 = 4

√3x + y = 8

∴ The equation of line in normal form is √3x + y = 8.

3. Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with the positive direction of x–axis is 15°.

Solution:

Given:

p = 4, α = 15°

The equation of the line in normal form is given by

We know that, cos 15° = cos (45° – 30°) = cos45°cos30° + sin45°sin30°

Cos (A – B) = cos A cos B + sin A sin B

So,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 25

And sin 15 = sin (45° – 30°) = sin 45° cos 30° – cos 45° sin 30°

Sin (A – B) = sin A cos B – cos A sin B

So,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 26

Now, by using the formula,

x cos α + y sin α = p

Now, substitute the values, we get

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 27

(√3+1)x +(√3-1) y = 8√2

∴ The equation of line in normal form is (√3+1)x +(√3-1) y = 8√2.

4. Find the equation of the straight line at a distance of 3 units from the origin such that the perpendicular from the origin to the line makes an angle α given by tan α = 5/12 with the positive direction of x–axis.

Solution:

Given:

p = 3, α = tan-1 (5/12)

So, tan α = 5/12

sin α = 5/13

cos α = 12/13

The equation of the line in normal form is given by

By using the formula,

x cos α + y sin α = p

Now, substitute the values, we get

12x/13 + 5y/13 = 3

12x + 5y = 39

∴ The equation of line in normal form is 12x + 5y = 39.

5. Find the equation of the straight line on which the length of the perpendicular from the origin is 2 and the perpendicular makes an angle α with x–axis such that sin α = 1/3.

Solution:

Given:

p = 2, sin α = 1/3

We know that cos α = √(1 – sin2 α)

= √(1 – 1/9)

= 2√2/3

The equation of the line in normal form is given by

By using the formula,

x cos α + y sin α = p

Now, substitute the values, we get

x2√2/3 + y/3 = 2

2√2x + y = 6

∴ The equation of line in normal form is 2√2x + y = 6.

EXERCISE 23.8 PAGE NO: 23.65

1. A line passes through a point A (1, 2) and makes an angle of 600 with the x–axis and intercepts the line x + y = 6 at the point P. Find AP.

Solution:

Given:

(x1, y1) = A (1, 2), θ = 60°

Let us find the distance AP.

By using the formula,

The equation of the line is given by:

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 28

Here, r represents the distance of any point on the line from point A (1, 2).

The coordinate of any point P on this line are (1 + r/2, 2 + √3r/2) 

It is clear that, P lies on the line x + y = 6

So,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 29

∴ The value of AP is 3(√3 – 1)

2. If the straight line through the point P(3, 4) makes an angle π/6 with the x–axis and meets the line 12x + 5y + 10 = 0 at Q, find the length PQ.

Solution:

Given:

(x1, y1) = A (3, 4), θ = π/6 = 30°

Let us find the length PQ.

By using the formula,

The equation of the line is given by:

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 30

 x – √3 y + 4√3 – 3 = 0

Let PQ = r

Then, the coordinate of Q are given by

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 31

3. A straight line drawn through the point A (2, 1) making an angle π/4 with positive x–axis intersects another line x + 2y + 1 = 0 in the point B. Find length AB.

Solution:

Given:

(x1, y1) = A (2, 1), θ = π/4 = 45°

Let us find the length AB.

By using the formula,

The equation of the line is given by:

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 32

x – y – 1 = 0

Let AB = r

Then, the coordinate of B is given by

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 33

4. A line a drawn through A (4, – 1) parallel to the line 3x – 4y + 1 = 0. Find the coordinates of the two points on this line which are at a distance of 5 units from A.

Solution:

Given:

(x1, y1) = A (4, -1)

Let us find Coordinates of the two points on this line which are at a distance of 5 units from A.

Given: Line 3x – 4y + 1 = 0

4y = 3x + 1

y = 3x/4 + 1/4

Slope tan θ = 3/4  

So,

Sin θ = 3/5

Cos θ = 4/5

The equation of the line passing through A (4, −1) and having slope ¾ is

By using the formula,

The equation of the line is given by:

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 34

3x – 4y = 16


Here, AP = r = ± 5

Thus, the coordinates of P are given by

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 35

x
https://gradeup-question-images.grdp.co/liveData/PROJ26150/1549601523027278.png= ±4 + 4 and y = ±3 –1

x = 8, 0 and y = 2, – 4

∴ The coordinates of the two points at a distance of 5 units from A are (8, 2) and (0, −4).

5. The straight line through P(x1, y1) inclined at an angle θ with the x–axis meets the line ax + by + c = 0 in Q. Find the length of PQ.

Solution:

Given:

The equation of the line that passes through P(x1, y1) and makes an angle of θ with the x–axis.

Let us find the length of PQ.

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 36

By using the formula,

The equation of the line is given by:

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 37

EXERCISE 23.9 PAGE NO: 23.72

1. Reduce the equation √3x + y + 2 = 0 to:
(i) slope – intercept form and find slope and y – intercept;
(ii) Intercept form and find intercept on the axes
(iii) The normal form and find p and α.

Solution:

(i) Given: 

√3x + y + 2 = 0 

y = – √3x – 2

This is the slope intercept form of the given line.

∴ The slope = – √3 and y – intercept = -2

(ii) Given: 

√3x + y + 2 = 0 

√3x + y = -2

Divide both sides by -2, we get

√3x/-2 + y/-2 = 1

∴ The intercept form of the given line. Here, x – intercept = – 2/√3 and y – intercept = -2

(iii) Given: 

√3x + y + 2 = 0 

-√3x – y = 2

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 38

2. Reduce the following equations to the normal form and find p and α in each case:
(i) x + √3y – 4 = 0
(ii) x + y + √2 = 0

Solution:

(i) x + √3y – 4 = 0

x + √3y = 4

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 39

The normal form of the given line, where p = 2, cos α = 1/2 and sin α = √3/2 

∴ p = 2 and α = π/3

(ii) x + y + √2 = 0

-x – y = √2

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 40

The normal form of the given line, where p = 1, cos α = -1/√2 and sin α = -1/√2

∴ p = 1 and α = 225o

3. Put the equation x/a + y/b = 1 the slope intercept form and find its slope and y – intercept.

Solution:

Given: the equation is x/a + y/b = 1 

We know that,

General equation of line y = mx + c.

bx + ay = ab

ay = – bx + ab

y = -bx/a + b

The slope intercept form of the given line.

∴ Slope = – b/a and y – intercept = b

4. Reduce the lines 3x – 4y + 4 = 0 and 2x + 4y – 5 = 0 to the normal form and hence find which line is nearer to the origin.

Solution:

Given:

The normal forms of the lines 3x − 4y + 4 = 0 and 2x + 4y − 5 = 0.

Let us find, in given normal form of a line, which is nearer to the origin.

-3x + 4y = 4

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 41

Now 2x + 4y = – 5

-2x – 4y = 5

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 42

From equations (1) and (2):

45 < 525

∴ The line 3x − 4y + 4 = 0 is nearer to the origin.

5. Show that the origin is equidistant from the lines 4x + 3y + 10 = 0; 5x – 12y + 26 = 0 and 7x + 24y = 50.

Solution:

Given:

The lines 4x + 3y + 10 = 0; 5x – 12y + 26 = 0 and 7x + 24y = 50.

We need to prove that, the origin is equidistant from the lines 4x + 3y + 10 = 0; 5x – 12y + 26 = 0 and 7x + 24y = 50.

Let us write down the normal forms of the given lines.

First line: 4x + 3y + 10 = 0

-4x – 3y = 10

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 43

So, p = 2

Second line: 5x − 12y + 26 = 0

-5x + 12y = 26

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 44

So, p = 2

Third line: 7x + 24y = 50

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 45

So, p = 2

∴ The origin is equidistant from the given lines.

EXERCISE 23.10 PAGE NO: 23.77

1. Find the point of intersection of the following pairs of lines:
(i) 2x – y + 3 = 0 and x + y – 5 = 0

(ii) bx + ay = ab and ax + by = ab

Solution:

(i) 2x – y + 3 = 0 and x + y – 5 = 0

Given:

The equations of the lines are as follows:

2x − y + 3 = 0 … (1)

x + y − 5 = 0 … (2)

Let us find the point of intersection of pair of lines.

By solving (1) and (2) using cross – multiplication method, we get

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 46

x = 2/3 and y = 13/3

∴ The point of intersection is (2/3, 13/3)

(ii) bx + ay = ab and ax + by = ab

Given:

The equations of the lines are as follows:

bx + ay − ab = 0… (1)

ax + by = ab ⇒ ax + by − ab = 0 … (2)

Let us find the point of intersection of pair of lines.

By solving (1) and (2) using cross – multiplication method, we get

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 47

∴ The point of intersection is (ab/a+b, ab/a+b)

2. Find the coordinates of the vertices of a triangle, the equations of whose sides are: (i) x + y – 4 = 0, 2x – y + 3 0 and x – 3y + 2 = 0

(ii) y (t1 + t2) = 2x + 2at1t2, y (t2 + t3) = 2x + 2at2t3 and, y(t3 + t1) = 2x + 2at1t3.

Solution:

(i) x + y – 4 = 0, 2x – y + 3 0 and x – 3y + 2 = 0

Given:

x + y − 4 = 0, 2x − y + 3 = 0 and x − 3y + 2 = 0

Let us find the point of intersection of pair of lines.

x + y − 4 = 0 … (1)

2x − y + 3 = 0 … (2)

x − 3y + 2 = 0 … (3)

By solving (1) and (2) using cross – multiplication method, we get

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 48

 x = 1/3, y = 11/3


Solving (1) and (3) using cross – multiplication method, we get

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 49

x = 5/2, y = 3/2


Similarly, solving (2) and (3) using cross – multiplication method, we get

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 50

x = – 7/5, y = 1/5

∴ The coordinates of the vertices of the triangle are (1/3, 11/3), (5/2, 3/2) and (-7/5, 1/5)

(ii) y (t1 + t2) = 2x + 2at1t2, y (t2 + t3) = 2x + 2at2t3 and, y(t3 + t1) = 2x + 2at1t3.

Given:

y (t1 + t2) = 2x + 2a t1t2, y (t2 + t3) = 2x + 2a t2t3 and y (t3 + t1) = 2x + 2a t1t3

Let us find the point of intersection of pair of lines.

2x − y (t1 + t2) + 2a t1t2 = 0 … (1)

2x − y (t2 + t3) + 2a t2t3 = 0 … (2)

2x − y (t3 + t1) + 2a t1t3 = 0 … (3)

By solving (1) and (2) using cross – multiplication method, we get

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 51

Solving (1) and (3) using cross – multiplication method, we get

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 52

Solving (2) and (3) using cross – multiplication method, we get

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 53

∴ The coordinates of the vertices of the triangle are (at21, 2at1), (at22, 2at2) and (at23, 2at3).

3. Find the area of the triangle formed by the lines
y = m1x + c1, y = m2x + c2 and x = 0

Solution:

Given:

y = m1x + c1 … (1)

y = m2x + c2 … (2)

x = 0 … (3)

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Solving (1) and (2), we get

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 54

Solving (1) and (3):

x = 0, y = c1

Thus, AB and CA intersect at A 0,c1.


Similarly, solving (2) and (3):

x = 0, y = c2

Thus, BC and CA intersect at C 0,c2.

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 55

4. Find the equations of the medians of a triangle, the equations of whose sides are:
3x + 2y + 6 = 0, 2x – 5y + 4 = 0 and x – 3y – 6 = 0

Solution:

Given:

3x + 2y + 6 = 0 … (1)

2x − 5y + 4 = 0 … (2)

x − 3y − 6 = 0 … (3)

Let us assume, in triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Solving equations (1) and (2), we get

x = −2, y = 0

Thus, AB and BC intersect at B (−2, 0).

Now, solving (1) and (3), we get

x = – 6/11, y = – 24/11

Thus, AB and CA intersect at A (-6/11, -24/11) 


Similarly, solving (2) and (3), we get

x = −42, y = −16

Thus, BC and CA intersect at C (−42, −16).


Now, let D, E and F be the midpoints the sides BC, CA and AB, respectively.

Then, we have:

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 56

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 57

∴ The equations of the medians of a triangle are: 41x – 112y – 70 = 0,

16x – 59y – 120 = 0, 25x – 53y + 50 = 0

5. Prove that the lines y = √3x + 1, y = 4 and y = -√3x + 2 form an equilateral triangle.

Solution:

Given:

y = √3x + 1…… (1)

y = 4 ……. (2)

y = – √3x + 2……. (3)

Let us assume in triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

By solving equations (1) and (2), we get

x = √3, y = 4

Thus, AB and BC intersect at B(√3,4)


Now, solving equations (1) and (3), we get

x = 1/2√3, y = 3/2

Thus, AB and CA intersect at A (1/2√3, 3/2)


Similarly, solving equations (2) and (3), we get

x = -2/√3, y = 4

Thus, BC and AC intersect at C (-2/√3,4)


Now, we have:

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 58

Hence proved, the given lines form an equilateral triangle.

EXERCISE 23.11 PAGE NO: 23.83

1. Prove that the following sets of three lines are concurrent:
(i) 15x – 18y + 1 = 0, 12x + 10y – 3 = 0 and 6x + 66y – 11 = 0

(ii) 3x – 5y – 11 = 0, 5x + 3y – 7 = 0 and x + 2y = 0

Solution:

(i) 15x – 18y + 1 = 0, 12x + 10y – 3 = 0 and 6x + 66y – 11 = 0

Given:

15x – 18y + 1 = 0 …… (i)

12x + 10y – 3 = 0 …… (ii)

6x + 66y – 11 = 0 …… (iii)

Now, consider the following determinant:

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 59

=> 1320 – 2052 + 732 = 0

Hence proved, the given lines are concurrent.

(ii) 3x – 5y – 11 = 0, 5x + 3y – 7 = 0 and x + 2y = 0

Given:

3x − 5y − 11 = 0 …… (i)

5x + 3y − 7 = 0 …… (ii)

x + 2y = 0 …… (iii)

Now, consider the following determinant:

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 60

Hence, the given lines are concurrent.

2. For what value of λ are the three lines 2x – 5y + 3 = 0, 5x – 9y + λ = 0 and x – 2y + 1 = 0 concurrent?

Solution:

Given:

2x − 5y + 3 = 0 … (1)

5x − 9y + λ = 0 … (2)

x − 2y + 1 = 0 … (3)

It is given that the three lines are concurrent.

Now, consider the following determinant:

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 61

2(-9 + 2λ) + 5(5 – λ) + 3(-10 + 9) = 0

-18 + 4λ + 25 – 5λ – 3 = 0

λ = 4

∴ The value of λ is 4.

3. Find the conditions that the straight lines y = m1x + c1, y = m2x + c2 and y = m3x + c3 may meet in a point.

Solution:

Given:

m1x – y + c1 = 0 … (1)

m2x – y + c2 = 0 … (2)

m3x – y + c3 = 0 … (3)

It is given that the three lines are concurrent.

Now, consider the following determinant:

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 62

m1(-c3 + c2) + 1(m2c3-m3c2) + c1(-m2 + m3) = 0

m1(c2-c3) + m2(c3-c1) + m3(c1-c2) = 0

∴ The required condition is m1(c2-c3) + m2(c3-c1) + m3(c1-c2) = 0

4. If the lines p1x + q1y = 1, p2x + q2y = 1 and p3x + q3y = 1 be concurrent, show that the points (p1, q1), (p2, q2) and (p3, q3) are collinear.

Solution:

Given:

p1x + q1y = 1

p2x + q2y = 1

p3x + q3y = 1

The given lines can be written as follows:

p1 x + q1 y – 1 = 0 … (1)

p2 x + q2 y – 1 = 0 … (2)

p3 x + q3 y – 1 = 0 … (3)

It is given that the three lines are concurrent.

Now, consider the following determinant:

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 63

Hence proved, the given three points, (p1, q1), (p2, q2) and (p3, q3) are collinear.

5. Show that the straight lines L1 = (b + c)x + ay + 1 = 0, L2 = (c + a)x + by + 1 = 0 and L3 = (a + b)x + cy + 1 = 0 are concurrent.

Solution:

Given:

L1 = (b + c)x + ay + 1 = 0

L2 = (c + a)x + by + 1 = 0

L3 = (a + b)x + cy + 1 = 0

The given lines can be written as follows:

(b + c) x + ay + 1 = 0 … (1)

(c + a) x + by + 1 = 0 … (2)

(a + b) x + cy + 1 = 0 … (3)

Consider the following determinant.

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 64

Hence proved, the given lines are concurrent.

EXERCISE 23.12 PAGE NO: 23.92

1. Find the equation of a line passing through the point (2, 3) and parallel to the line 3x– 4y + 5 = 0.

Solution:

Given: 

The equation is parallel to 3x − 4y + 5 = 0 and pass through (2, 3)

The equation of the line parallel to 3x − 4y + 5 = 0 is

3x – 4y + λ = 0,

Where, λ is a constant.

It passes through (2, 3).

Substitute the values in above equation, we get

3 (2) – 4 (3) + λ = 0

6 – 12 + λ = 0

λ = 6

Now, substitute the value of λ = 6 in 3x – 4y + λ = 0, we get

3x − 4y + 6

∴ The required line is 3x − 4y + 6 = 0.

2. Find the equation of a line passing through (3, -2) and perpendicular to the line x – 3y + 5 = 0.

Solution:

Given: 

The equation is perpendicular to x – 3y + 5 = 0 and passes through (3,-2)

The equation of the line perpendicular to x − 3y + 5 = 0 is

3x + y + λ = 0,

Where, λ is a constant.

It passes through (3, − 2).

Substitute the values in above equation, we get

3 (3) + (-2) + λ = 0

9 – 2 + λ = 0

λ = – 7

Now, substitute the value of λ = − 7 in 3x + y + λ = 0, we get

3x + y – 7 = 0

∴ The required line is 3x + y – 7 = 0.

3. Find the equation of the perpendicular bisector of the line joining the points (1, 3) and (3, 1).

Solution:

Given: 

A (1, 3) and B (3, 1) be the points joining the perpendicular bisector

Let C be the midpoint of AB.

So, coordinates of C = [(1+3)/2, (3+1)/2] 

= (2, 2)

Slope of AB = [(1-3) / (3-1)] 

= -1

Slope of the perpendicular bisector of AB = 1

Thus, the equation of the perpendicular bisector of AB is given as,

y – 2 = 1(x – 2)

y = x

x – y = 0

∴ The required equation is y = x.

4. Find the equations of the altitudes of a ΔABC whose vertices are A (1, 4), B (-3, 2) and C (-5, -3).

Solution:

Given: 

The vertices of ∆ABC are A (1, 4), B (− 3, 2) and C (− 5, − 3).

Now let us find the slopes of ∆ABC.

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 65

Slope of AB = [(2 – 4) / (-3-1)] 

= ½

Slope of BC = [(-3 – 2) / (-5+3)] 

= 5/2  


Slope of CA = [(4 + 3) / (1 + 5)] 

= 7/6 

Thus, we have:

Slope of CF = -2

Slope of AD = -2/5 

Slope of BE = -6/7 

Hence,

Equation of CF is:

y + 3 = -2(x + 5)

y + 3 = -2x – 10

2x + y + 13 = 0

Equation of AD is:

y – 4 = (-2/5) (x – 1)

5y – 20 = -2x + 2

2x + 5y – 22 = 0


Equation of BE is:

y – 2 = (-6/7) (x + 3)

7y – 14 = -6x – 18

6x + 7y + 4 = 0

∴ The required equations are 2x + y + 13 = 0, 2x + 5y – 22 = 0, 6x + 7y + 4 = 0.

5. Find the equation of a line which is perpendicular to the line√3x – y + 5 = 0 and which cuts off an intercept of 4 units with the negative direction of y-axis.

Solution:

Given: 

The equation is perpendicular to √3x – y + 5 = 0 equation and cuts off an intercept of 4 units with the negative direction of y-axis.

The line perpendicular to √3x – y + 5 = 0 is x + √3y + λ = 0  

It is given that the line x + √3y + λ = 0 cuts off an intercept of 4 units with the negative direction of the y-axis.

This means that the line passes through (0,-4).

So,

Let us substitute the values in the equation x + √3y + λ = 0, we get

0 – √3 (4) + λ = 0 

λ = 4√3

Now, substitute the value of λ back, we get

x + √3y + 4√3 = 0

∴ The required equation of line is x + √3y + 4√3 = 0.

EXERCISE 23.13 PAGE NO: 23.99

1. Find the angles between each of the following pairs of straight lines:
(i) 3x + y + 12 = 0 and x + 2y – 1 = 0

(ii) 3x – y + 5 = 0 and x – 3y + 1 = 0

Solution:

(i) 3x + y + 12 = 0 and x + 2y – 1 = 0

Given:

The equations of the lines are

3x + y + 12 = 0 … (1)

x + 2y − 1 = 0 … (2)

Let m1 and m2 be the slopes of these lines.

m1 = -3, m2 = -1/2

Let θ be the angle between the lines.

Then, by using the formula

tan θ = [(m1 – m2) / (1 + m1m2)]

= [(-3 + 1/2) / (1 + 3/2)]

= 1

So,

θ = π/4 or 45o

∴ The acute angle between the lines is 45°

(ii) 3x – y + 5 = 0 and x – 3y + 1 = 0

Given:

The equations of the lines are

3x − y + 5 = 0 … (1)

x − 3y + 1 = 0 … (2)

Let m1 and m2 be the slopes of these lines.

m1 = 3, m2 = 1/3

Let θ be the angle between the lines.

Then, by using the formula

tan θ = [(m1 – m2) / (1 + m1m2)]

= [(3 + 1/3) / (1 + 1)]

= 4/3

So,

θ = tan-1 (4/3)

∴ The acute angle between the lines is tan-1 (4/3).

2. Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0.

Solution:

Given:

The equations of the lines are

2x − y + 3 = 0 … (1)

x + y + 2 = 0 … (2)

Let m1 and m2 be the slopes of these lines.

m1 = 2, m2 = -1

Let θ be the angle between the lines.

Then, by using the formula

tan θ = [(m1 – m2) / (1 + m1m2)]

= [(2 + 1) / (1 + 2)]

= 3

So,

θ = tan-1 (3)

∴ The acute angle between the lines is tan-1 (3).

3. Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.

Solution:

To prove:

The points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram

Let us assume the points, A (2, − 1), B (0, 2), C (2, 3) and D (4, 0) be the vertices.

Now, let us find the slopes

Slope of AB = [(2+1) / (0-2)]

= -3/2

Slope of BC = [(3-2) / (2-0)]

= ½

Slope of CD = [(0-3) / (4-2)]

= -3/2

Slope of DA = [(-1-0) / (2-4)]

= ½

Thus, AB is parallel to CD and BC is parallel to DA.

Hence proved, the given points are the vertices of a parallelogram.


Now, let us find the angle between the diagonals AC and BD.

Let m1 and m2 be the slopes of AC and BD, respectively.

m1 = [(3+1) / (2-2)]

= ∞

m2 = [(0-2) / (4-0)]

= -1/2

Thus, the diagonal AC is parallel to the y-axis.

∠ODB = tan-1 (1/2)


In triangle MND,

∠DMN = π/2 – tan-1 (1/2)

∴ The angle between the diagonals is π/2 – tan-1 (1/2).

4. Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.

Solution:

Given:

Points (2, 0), (0, 3) and the line x + y = 1.

Let us assume A (2, 0), B (0, 3) be the given points.

Now, let us find the slopes

Slope of AB = m1

= [(3-0) / (0-2)]

= -3/2

Slope of the line x + y = 1 is -1

∴ m2 = -1


Let θ be the angle between the line joining the points (2, 0), (0, 3) and the line x + y =

tan θ = |[(m1 – m2) / (1 + m1m2)]|

= [(-3/2 + 1) / (1 + 3/2)]

= 1/5

θ = tan-1 (1/5)

∴ The acute angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1 is tan-1 (1/5).

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 66

Solution:

We need to prove:

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 67

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 68

Let us assume A (x1, y1) and B (x2, y2) be the given points and O be the origin.

Slope of OA = m1 = y1×1

Slope of OB = m2 = y2×2

It is given that θ is the angle between lines OA and OB.

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 69

Hence proved.

EXERCISE 23.14 PAGE NO: 23.102

1. Find the values of α so that the point P(α 2, α) lies inside or on the triangle formed by the lines x – 5y + 6 = 0, x – 3y + 2 = 0 and x – 2y – 3 = 0.

Solution:

Given: 

x – 5y + 6 = 0, x – 3y + 2 = 0 and x – 2y – 3 = 0 forming a triangle and point P(α2, α) lies inside or on the triangle

Let ABC be the triangle of sides AB, BC and CA whose equations are x − 5y + 6 = 0, x − 3y + 2 = 0 and x − 2y − 3 = 0, respectively.

On solving the equations, we get A (9, 3), B (4, 2) and C (13, 5) as the coordinates of the vertices.

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 70

It is given that point P (α2, α) lies either inside or on the triangle. The three conditions are given below.

(i) A and P must lie on the same side of BC.

(ii) B and P must lie on the same side of AC.

(iii) C and P must lie on the same side of AB.


If A and P lie on the same side of BC, then

(9 – 9 + 2)(α2 – 3α + 2) ≥0

(α – 2)(α – 1) ≥ 0

α ∈ (- ∞, 1 ] ∪ [ 2, ∞) … (1)


If B and P lie on the same side of AC, then

(4 – 4 – 3) (α2 – 2α – 3) ≥ 0

(α – 3)(α + 1) ≤ 0

α ∈ [- 1, 3] … (2)


If C and P lie on the same side of AB, then

(13 – 25 + 6)(α2 – 5α + 6) ≥0

(α – 3)(α – 2) ≤ 0

α ∈ [ 2, 3] … (3)


From equations (1), (2) and (3), we get

α∈ [2, 3]

∴ α∈ [2, 3]

2. Find the values of the parameter a so that the point (a, 2) is an interior point of the triangle formed by the lines x + y – 4 = 0, 3x – 7y – 8 = 0 and 4x – y – 31 = 0.

Solutions:

Given:

x + y – 4 = 0, 3x – 7y – 8 = 0 and 4x – y – 31 = 0 forming a triangle and point (a, 2)is an interior point of the triangle

Let ABC be the triangle of sides AB, BC and CA whose equations are x + y − 4 = 0, 3x − 7y − 8 = 0 and 4x − y − 31 = 0, respectively.

On solving them, we get A (7, – 3), B (18/5, 2/5) and C (209/25, 61/25) as the coordinates of the vertices.

Let P (a, 2) be the given point.

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 71

It is given that point P (a, 2) lies inside the triangle. So, we have the following:

(i) A and P must lie on the same side of BC.

(ii) B and P must lie on the same side of AC.

(iii) C and P must lie on the same side of AB.

Thus, if A and P lie on the same side of BC, then

21 + 21 – 8 – 3a – 14 – 8 > 0

a > 22/3 … (1)

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 72

From (1), (2) and (3), we get:

A ∈ (22/3, 33/4) 

∴ A ∈ (22/3, 33/4) 

3. Determine whether the point (-3, 2) lies inside or outside the triangle whose sides are given by the equations x + y – 4 = 0, 3x – 7y + 8 = 0, 4x – y – 31 = 0.

Solution:

Given:

x + y – 4 = 0, 3x – 7y + 8 = 0, 4x – y – 31 = 0 forming a triangle and point (-3, 2)

Let ABC be the triangle of sides AB, BC and CA, whose equations x + y − 4 = 0, 3x − 7y + 8 = 0 and 4x − y − 31 = 0, respectively.

On solving them, we get A (7, – 3), B (2, 2) and C (9, 5) as the coordinates of the vertices.

Let P (− 3, 2) be the given point.

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 73

The given point P (− 3, 2) will lie inside the triangle ABC, if

(i) A and P lies on the same side of BC

(ii) B and P lies on the same side of AC

(iii) C and P lies on the same side of AB


Thus, if A and P lie on the same side of BC, then

21 + 21 + 8 – 9 – 14 + 8 > 0

50 × – 15 > 0

-750 > 0,

This is false

∴ The point (−3, 2) lies outside triangle ABC.

EXERCISE 23.15 PAGE NO: 23.107

1. Find the distance of the point (4, 5) from the straight line 3x – 5y + 7 = 0.

Solution:

Given:

The line: 3x – 5y + 7 = 0

Comparing ax + by + c = 0 and 3x − 5y + 7 = 0, we get:

a = 3, b = − 5 and c = 7

So, the distance of the point (4, 5) from the straight line 3x − 5y + 7 = 0 is

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 74

∴ The required distance is 6/34

2. Find the perpendicular distance of the line joining the points (cos θ, sin θ) and (cos ϕ, sin ϕ) from the origin.

Solution:

Given: 

The points (cos θ, sin θ) and (cos ϕ, sin ϕ) from the origin.

The equation of the line joining the points (cos θ, sin θ) and (cos ϕ, sin ϕ) is given below:

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 75

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 76

3. Find the length of the perpendicular from the origin to the straight line joining the two points whose coordinates are (a cos α, a sin α) and (a cos β, a sin β).

Solution:

Given:

Coordinates are (a cos α, a sin α) and (a cos β, a sin β).

Equation of the line passing through (a cos α, a sin α) and (a cos β, a sin β) is

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 77

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 78

4. Show that the perpendicular let fall from any point on the straight line 2x + 11y – 5 = 0 upon the two straight lines 24x + 7y = 20 and 4x – 3y – 2 = 0 are equal to each other.

Solution:

Given:

The lines 24x + 7y = 20 and 4x – 3y – 2 = 0

Let us assume, P(a, b) be any point on 2x + 11y − 5 = 0

So,

2a + 11b − 5 = 0

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 79

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 80

5. Find the distance of the point of intersection of the lines 2x + 3y = 21 and 3x – 4y + 11 = 0 from the line 8x + 6y + 5 = 0.

Solution:

Given:

The lines 2x + 3y = 21 and 3x – 4y + 11 = 0

Solving the lines 2x + 3y = 21 and 3x − 4y + 11 = 0 we get:

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 81

x = 3, y = 5

So, the point of intersection of 2x + 3y = 21 and 3x − 4y + 11 = 0 is (3, 5).

Now, the perpendicular distance d of the line 8x + 6y + 5 = 0 from the point (3, 5) is

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 82

∴ The distance is 59/10.

EXERCISE 23.16 PAGE NO: 23.114

1. Determine the distance between the following pair of parallel lines:
(i) 4x – 3y – 9 = 0 and 4x – 3y – 24 = 0

(ii) 8x + 15y – 34 = 0 and 8x + 15y + 31 = 0

Solution:

(i) 4x – 3y – 9 = 0 and 4x – 3y – 24 = 0

Given: 

The parallel lines are

4x − 3y − 9 = 0 … (1)

4x − 3y − 24 = 0 … (2)

Let d be the distance between the given lines.

So,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 83

∴ The distance between givens parallel line is 3units.

(ii) 8x + 15y – 34 = 0 and 8x + 15y + 31 = 0

Given: 

The parallel lines are

8x + 15y − 34 = 0 … (1)

8x + 15y + 31 = 0 … (2)

Let d be the distance between the given lines.

So,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 84

∴ The distance between givens parallel line is 65/17 units.

2. The equations of two sides of a square are 5x – 12y – 65 = 0 and 5x – 12y + 26 = 0. Find the area of the square.

Solution:

Given: 

Two side of square are 5x – 12y – 65 = 0 and 5x – 12y + 26 = 0

The sides of a square are

5x − 12y − 65 = 0 … (1)

5x − 12y + 26 = 0 … (2)

We observe that lines (1) and (2) are parallel.

So, the distance between them will give the length of the side of the square.

Let d be the distance between the given lines.

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 85

∴ Area of the square = 72 = 49 square units

3. Find the equation of two straight lines which are parallel to x + 7y + 2 = 0 and at unit distance from the point (1, -1).

Solution:

Given: 

The equation is parallel to x + 7y + 2 = 0 and at unit distance from the point (1, -1)

The equation of given line is

x + 7y + 2 = 0 … (1)

The equation of a line parallel to line x + 7y + 2 = 0 is given below:

x + 7y + λ = 0 … (2)

The line x + 7y + λ = 0 is at a unit distance from the point (1, − 1).

So,

1 = 1 – 7 + λ1 + 49

λ – 6 = ± 52

λ = 6 + 52, 6 – 52

now, substitute the value of λ back in equation x + 7y + λ = 0, we get

x + 7y + 6 + 52 = 0 and x + 7y + 6 – 52

∴ The required lines:

x + 7y + 6 + 52 = 0 and x + 7y + 6 – 52

4. Prove that the lines 2x + 3y = 19 and 2x + 3y + 7 = 0 are equidistant from the line 2x + 3y = 6.

Solution:

Given: 

The lines A, 2x + 3y = 19 and B, 2x + 3y + 7 = 0 also a line C, 2x + 3y = 6.

Let d1 be the distance between lines 2x + 3y = 19 and 2x + 3y = 6,

While d2 is the distance between lines 2x + 3y + 7 = 0 and 2x + 3y = 6

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 86

Hence proved, the lines 2x + 3y = 19 and 2x + 3y + 7 = 0 are equidistant from the line 2x + 3y = 6

5. Find the equation of the line mid-way between the parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.

Solution:

Given:

9x + 6y – 7 = 0 and 3x + 2y + 6 = 0 are parallel lines

The given equations of the lines can be written as:

3x + 2y – 7/3 = 0 … (1)

3x + 2y + 6 = 0 … (2)

Let the equation of the line midway between the parallel lines (1) and (2) be

3x + 2y + λ = 0 … (3)

The distance between (1) and (3) and the distance between (2) and (3) are equal.

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 87

Now substitute the value of λ back in equation 3x + 2y + λ = 0, we get

3x + 2y + 11/6 = 0

By taking LCM

18x + 12y + 11 = 0

∴ The required equation of line is 18x + 12y + 11 = 0

EXERCISE 23.17 PAGE NO: 23.117

1. Prove that the area of the parallelogram formed by the lines
a1x + b1y + c1 = 0, a1x + b1y + d1 = 0, a2x + b2y + c2 = 0, a2x + b2y + d2 = 0 is RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 88 sq. units.
Deduce the condition for these lines to form a rhombus.

Solution:

Given:

The given lines are

a1x + b1y + c1 = 0 … (1)

a1x + b1y + d1 = 0 … (2)

a2x + b2y + c2 = 0 … (3)

a2x + b2y + d2 = 0 … (4)

Let us prove, the area of the parallelogram formed by the lines a1x + b1y + c1 = 0, a1x + b1y + d1 = 0, a2x + b2y + c2 = 0, a2x + b2y + d2 = 0 is 
RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 89 sq. units.

The area of the parallelogram formed by the lines a1x + b1y + c1 = 0, a1x + b1y + d1 = 0, a2x + b2y + c2 = 0 and a2x + b2y + d2 = 0 is given below:

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 90

Hence proved.

2. Prove that the area of the parallelogram formed by the lines 3x – 4y + a = 0, 3x –4y + 3a = 0, 4x – 3y – a = 0 and 4x – 3y – 2a = 0 is 2a2/7 sq. units.

Solution:

Given:

The given lines are

3x − 4y + a = 0 … (1)

3x − 4y + 3a = 0 … (2)

4x − 3y − a = 0 … (3)

4x − 3y − 2a = 0 … (4)

Let us prove, the area of the parallelogram formed by the lines 3x – 4y + a = 0, 3x – 4y + 3a = 0, 4x – 3y – a = 0 and 4x – 3y – 2a = 0 is 2a2/7 sq. units.

From above solution, we know that

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 91

Hence proved.

3. Show that the diagonals of the parallelogram whose sides are lx + my + n = 0, lx + my + n’ = 0, mx + ly + n = 0 and mx + ly + n’ = 0 include an angle π/2.

Solution:

Given:

The given lines are

lx + my + n = 0 … (1)

mx + ly + n’ = 0 … (2)

lx + my + n’ = 0 … (3)

mx + ly + n = 0 … (4)

Let us prove, the diagonals of the parallelogram whose sides are lx + my + n = 0, lx + my + n’ = 0, mx + ly + n = 0 and mx + ly + n’ = 0 include an angle π/2.

By solving (1) and (2), we get

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 92

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 93

∴ m1m2 = -1

Hence proved.

EXERCISE 23.18 PAGE NO: 23.124

1. Find the equation of the straight lines passing through the origin and making an angle of 45o with the straight line √3x + y = 11.

Solution:

Given:

Equation passes through (0, 0) and make an angle of 45° with the line √3x + y = 11.

We know that, the equations of two lines passing through a point x1,y1 and making an angle α with the given line y = mx + c are

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 94

2. Find the equations to the straight lines which pass through the origin and are inclined at an angle of 75o to the straight line x + y + √3(y – x) = a.

Solution:

Given:

The equation passes through (0,0) and make an angle of 75° with the line x + y + √3(y – x) = a.

We know that the equations of two lines passing through a point x1,y1 and making an angle α with the given line y = mx + c are

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 95

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 96

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 97

3. Find the equations of straight lines passing through (2, -1) and making an angle of 45o with the line 6x + 5y – 8 = 0.

Solution:

Given: 

The equation passes through (2,-1) and make an angle of 45° with the line 6x + 5y – 8 = 0

We know that the equations of two lines passing through a point x1, y1 and making an angle α with the given line y = mx + c are

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 98

Here, equation of the given line is,

6x + 5y – 8 = 0

5y = – 6x + 8

y = -6x/5 + 8/5

Comparing this equation with y = mx + c

We get, m = -6/5 

Where, x1 = 2, y1 = – 1, α = 45°, m = -6/5

So, the equations of the required lines are

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 99

x + 11y + 9 = 0 and 11x – y – 23 = 0

∴ The equation of given line is x + 11y + 9 = 0 and 11x – y – 23 = 0

4. Find the equations to the straight lines which pass through the point (h, k) and are inclined at angle tan-1 m to the straight line y = mx + c.

Solution:

Given: 

The equation passes through (h, k) and make an angle of tan-1 m with the line y = mx + c

We know that the equations of two lines passing through a point x1, y1 and making an angle α with the given line y = mx + c are

m′ = m

So,

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 100

Here,

x1 = h, y1 = k, α = tan-1 m, m′ = m.

So, the equations of the required lines are

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 101

5. Find the equations to the straight lines passing through the point (2, 3) and inclined at an angle of 450 to the lines 3x + y – 5 = 0.

Solution:

Given: 

The equation passes through (2, 3) and make an angle of 450with the line 3x + y – 5 = 0.

We know that the equations of two lines passing through a point x1,y1 and making an angle α with the given line y = mx + c are

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 102

Here,

Equation of the given line is,

3x + y – 5 = 0

y = – 3x + 5

Comparing this equation with y = mx + c we get, m = – 3

x1 = 2, y1 = 3, α = 45∘, m = – 3.

So, the equations of the required lines are

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 103

x + 2y – 8 = 0 and 2x – y – 1 = 0

∴ The equation of given line is x + 2y – 8 = 0 and 2x – y – 1 = 0

EXERCISE 23.19 PAGE NO: 23.124

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:

Given:

Lines 4x – 3y = 0 and 2x – 5y + 3 = 0 and parallel to 4x + 5 y + 6 = 0

The equation of the straight line passing through the points of intersection of 4x − 3y = 0 and 2x − 5y + 3 = 0 is given below:

4x − 3y + λ (2x − 5y + 3) = 0

(4 + 2λ)x + (− 3 − 5λ)y + 3λ = 0

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 104

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

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 105 

λ = -16/15

∴ The required equation is

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 106

28x + 35y – 48 = 0

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:

Given:

x + 2y + 3 = 0 and 3x + 4y + 7 = 0

The equation of the straight line passing through the points of intersection of x + 2y + 3 = 0 and 3x + 4y + 7 = 0 is

x + 2y + 3 + λ(3x + 4y + 7) = 0

(1 + 3λ)x + (2 + 4λ)y + 3 + 7λ = 0

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 107

The required line is perpendicular to x − y + 9 = 0 or, y = x + 9

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:

Given:

The equations, 2x – 7y + 11 = 0 and x + 3y – 8 = 0

The equation of the straight line passing through the points of intersection of 2x − 7y + 11 = 0 and x + 3y − 8 = 0 is given below:

2x − 7y + 11 + λ(x + 3y − 8) = 0

(2 + λ)x + (− 7 + 3λ)y + 11 − 8λ = 0


(i) The required line is parallel to the x-axis. So, the coefficient of x should be zero.

2 + λ = 0

λ = -2

Now, substitute the value of λ back in equation, we get

0 + (− 7 − 6)y + 11 + 16 = 0

13y − 27 = 0

∴ The equation of the required line is 13y − 27 = 0


(ii) The required line is parallel to the y-axis. So, the coefficient of y should be zero.

-7 + 3λ = 0

λ = 7/3

Now, substitute the value of λ back in equation, we get

(2 + 7/3)x + 0 + 11 – 8(7/3) = 0

13x – 23 = 0

∴ The equation of the required line is 13x – 23 = 0

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:

Given:

The equations, 2x + 3y + 1 = 0 and 3x – 5y – 5 = 0

The equation of the straight line passing through the points of intersection of 2x + 3y + 1 = 0 and 3x − 5y − 5 = 0 is

2x + 3y + 1 + λ(3x − 5y − 5) = 0

(2 + 3λ)x + (3 − 5λ)y + 1 − 5λ = 0

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

The required line is equally inclined to the axes. So, the slope of the required line is either 1 or − 1.

So,

– [(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, substitute the values of λ in (2 + 3λ)x + (3 − 5λ)y + 1 − 5λ = 0, we get the equations of the required lines as:

(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

∴ The required equation is 19x – 19y – 23 = 0 and 19x + 19y + 3 = 0

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:

Given:

The lines x + y = 4 and 2x – 3y = 1

The equation of the straight line passing through the point of intersection of x + y = 4 and 2x − 3y = 1 is

x + y − 4 + λ(2x − 3y − 1) = 0

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

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

The equation of the line with intercepts 5 and 6 on the axis is

x/5 + y/6 = 1 …. (2)

So, the slope of this line is -6/5

The lines (1) and (2) are perpendicular.

∴ -6/5 × [(-1+2λ) / (1 – 3λ)] = -1

λ = 11/3

Now, substitute the values of λ in (1), we get the equation of the required line.

(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

∴ The required equation is 25x – 30y – 23 = 0


Also, access exercises of RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines

Exercise 23.1 Solutions

Exercise 23.2 Solutions

Exercise 23.3 Solutions

Exercise 23.4 Solutions

Exercise 23.5 Solutions

Exercise 23.6 Solutions

Exercise 23.7 Solutions

Exercise 23.8 Solutions

Exercise 23.9 Solutions

Exercise 23.10 Solutions

Exercise 23.11 Solutions

Exercise 23.12 Solutions

Exercise 23.13 Solutions

Exercise 23.14 Solutions

Exercise 23.15 Solutions

Exercise 23.16 Solutions

Exercise 23.17 Solutions

Exercise 23.18 Solutions

Exercise 23.19 Solutions

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