RD Sharma Solutions for Class 11 Chapter 23 – Straight Lines

The RD Sharma Solutions for Class 11 Chapter 23 – Straight Lines is a crucial resource for students aiming to master the concepts of coordinate geometry. This chapter lays a solid foundation for higher-level mathematics, making it indispensable for those preparing for competitive exams like JEE. With clear explanations, step-by-step solutions, and diverse practice questions, this resource helps students build confidence and proficiency in understanding and applying straight line concepts.

The study of straight lines is fundamental to coordinate geometry and serves as the base for understanding more advanced geometric concepts. This chapter introduces key topics such as different forms of line equations, calculating slopes, and determining angles and distances between lines. Mastering straight lines is essential for progressing to advanced topics like circles, conic sections, and 3D geometry. The skills you acquire—like deriving line equations, checking parallelism and perpendicularity, and solving real-world problems—are crucial not just for exams but also for applications in fields like engineering, physics, computer graphics, and more.

What You Will Learn in RD Sharma Solutions for Class 11 Chapter 23 – Straight Lines

In this chapter, you will explore the following core concepts:

• Equations of Straight Lines: Learn the different forms of equations (e.g., slope-intercept, point-slope, etc.).
• Slope of a Line: Understand how to calculate the slope and its significance in geometry.
• Distance Between Two Parallel Lines: Master the formula and technique to find distances.
• Angle Between Two Lines: Learn to compute the angle between intersecting lines using the slope formula.
• Conditions for Parallelism and Perpendicularity: Derive equations and identify conditions for lines to be parallel or perpendicular.
• Applications of Straight Lines in Real Life: Understand how straight lines are used in fields like architecture, navigation, and physics.

Each concept is explained in detail with worked-out examples, making it easy for students to apply the formulas and solve problems confidently.

Do Check: RD Sharma Solutions for Class 6 to 12

The RD Sharma Solutions for Class 11 Chapter 23 offer detailed, step-by-step answers for all problems in the textbook. Written in simple and easy-to-understand language, RD Sharma Solutions for Class 11 ensure that even complex topics are easy to follow. The chapter also includes useful diagrams, solved examples, and practical tips to enhance your preparation.

Download RD Sharma Solutions for Class 11 Chapter 23 – PDF

The RD Sharma Class 11 Chapter 23 PDF contains comprehensive solutions, solved examples, and extra practice questions. Download the RD Sharma Solutions PDF to access all the material you need to build a solid understanding of straight lines and practice the concepts effectively.

What to Study Before Chapter 23 – Straight Lines

Before starting straight line chapter, ensure that you are familiar with basic concepts of the RD Sharma Solution Cartesian coordinate system, algebra, and plotting points on the coordinate plane. Reviewing the distance formula and section formula from the previous chapter will also help strengthen your foundation and make learning this chapter smoother.

RD Sharma Class 11 Chapter 23 – Exercises

Straight Line chapter includes a variety of exercises designed to reinforce and practice the concepts learned. Completing these exercises will help you gain a deeper understanding of the topic and improve problem-solving skills.

Question 1: Find the slope of the line passing through the points (2, 3) and (4, 7).

Solution:

Using the slope formula: m = (y₂ - y₁)/(x₂ - x₁)

Given points: (x₁, y₁) = (2, 3) and (x₂, y₂) = (4, 7)

m = (7 - 3)/(4 - 2)

m = 4/2

m = 2

Answer: Slope = 2

Question 2: Determine the equation of the line passing through (1, 2) with slope 3.

Solution:

Using point-slope form: y - y₁ = m(x - x₁)

Given: Point (1, 2) and slope m = 3

y - 2 = 3(x - 1)

y - 2 = 3x - 3

y = 3x - 3 + 2

y = 3x - 1

Answer: y = 3x - 1

Question 3: Write the slope-intercept form of the line whose x-intercept is 4 and y-intercept is –2.

Solution:

x-intercept = 4 means the line passes through (4, 0)

y-intercept = -2 means the line passes through (0, -2)

Using slope formula: m = (-2 - 0)/(0 - 4) = -2/(-4) = 1/2

Using slope-intercept form: y = mx + c

Where c = y-intercept = -2 and m = 1/2

y = (1/2)x + (-2)

y = (1/2)x - 2

Answer: y = (1/2)x - 2

Question 4: Find the angle between the lines y = 2x + 1 and y = -3x + 4.

Solution:

For lines with slopes m₁ and m₂, the acute angle θ between them is:

tan θ = |(m₁ - m₂)/(1 + m₁m₂)|

From the given equations: m₁ = 2 and m₂ = -3

tan θ = |2 - (-3)|/(1 + 2(-3))|

tan θ = |2 + 3|/(1 - 6)|

tan θ = |5|/|-5|

tan θ = 5/5 = 1

θ = tan⁻¹(1) = 45°

Answer: 45°

Question 5: Show that the points (1, 2), (4, 5), and (7, 8) lie on a straight line.

Solution:

To prove collinearity, we need to show that the slope between any two pairs of points is the same.

Slope between (1, 2) and (4, 5):

m₁ = (5 - 2)/(4 - 1) = 3/3 = 1

Slope between (4, 5) and (7, 8):

m₂ = (8 - 5)/(7 - 4) = 3/3 = 1

Since m₁ = m₂ = 1, the points are collinear.

Answer: The points lie on a straight line (proved)

Question 6: Find the equation of the line parallel to 3x + 4y = 7 and passing through (2, –1).

Solution:

Parallel lines have the same slope.

Rewriting 3x + 4y = 7 in slope-intercept form:

4y = -3x + 7

y = (-3/4)x + 7/4

Slope = -3/4

Using point-slope form with point (2, -1) and slope -3/4:

y - (-1) = (-3/4)(x - 2)

y + 1 = (-3/4)x + 3/2

y = (-3/4)x + 3/2 - 1

y = (-3/4)x + 1/2

Multiplying by 4: 4y = -3x + 2

3x + 4y = 2

Answer: 3x + 4y = 2

Question 7: Find the equation of the line perpendicular to 2x - 5y = 6 and passing through (0, 3).

Solution:

Rewriting 2x - 5y = 6 in slope-intercept form:

-5y = -2x + 6

y = (2/5)x - 6/5

Slope of given line = 2/5

Slope of perpendicular line = -1/(2/5) = -5/2

Using point-slope form with point (0, 3) and slope -5/2:

y - 3 = (-5/2)(x - 0)

y - 3 = (-5/2)x

y = (-5/2)x + 3

Multiplying by 2: 2y = -5x + 6

5x + 2y = 6

Answer: 5x + 2y = 6

Question 8: Find the distance between the parallel lines 2x + 3y = 4 and 2x + 3y = 10.

Solution:

For parallel lines ax + by = c₁ and ax + by = c₂, the distance is:

d = |c₁ - c₂|/√(a² + b²)

Here: a = 2, b = 3, c₁ = 4, c₂ = 10

d = |4 - 10|/√(2² + 3²)

d = |-6|/√(4 + 9)

d = 6/√13

d = 6√13/13 units

Answer: 6√13/13 units

Question 9: Determine the value of k if the line 3x + ky = 4 is parallel to the line 6x + 8y = 5.

Solution:

For two lines to be parallel, the ratios of coefficients of x and y must be equal.

For lines a₁x + b₁y = c₁ and a₂x + b₂y = c₂ to be parallel:

a₁/a₂ = b₁/b₂

Here: 3/6 = k/8

1/2 = k/8

k = 8/2 = 4

Answer: k = 4

Question 10: Find the equation of the line which is equidistant from the lines x + 2y = 7 and 3x + 6y = 8.

Solution:

First, rewrite the second equation in the same form as the first:

3x + 6y = 8 can be written as x + 2y = 8/3

These are parallel lines of the form x + 2y = c

The line equidistant from them has c = (7 + 8/3)/2

c = (21/3 + 8/3)/2 = (29/3)/2 = 29/6

Therefore, the equation is: x + 2y = 29/6

Multiplying by 6: 6x + 12y = 29

Answer: 6x + 12y = 29

Question 11: Find the coordinates of the foot of the perpendicular from the point (2, 3) to the line 4x - 3y + 8 = 0.

Solution:

Let the foot of perpendicular be (h, k)

The line from (2, 3) to (h, k) is perpendicular to the given line 4x - 3y + 8 = 0

Slope of given line = 4/3

Slope of perpendicular = -3/4

Slope from (2, 3) to (h, k) = (k - 3)/(h - 2) = -3/4

4(k - 3) = -3(h - 2)

4k - 12 = -3h + 6

3h + 4k = 18 ... (1)

Also, (h, k) lies on the given line:

4h - 3k + 8 = 0 ... (2)

From equation (2): 4h = 3k - 8

Substituting in equation (1): 3(3k - 8)/4 + 4k = 18

(9k - 24)/4 + 4k = 18

9k - 24 + 16k = 72

25k = 96

k = 96/25

h = (3 × 96/25 - 8)/4 = (288/25 - 200/25)/4 = 88/100 = 22/25

Answer: (22/25, 96/25)

Question 12: Find the equation of the line making equal intercepts on the axes and passing through (2, 3).

Solution:

If a line makes equal intercepts a on both axes, its equation is:

x + y = a

Since the line passes through (2, 3):

2 + 3 = a

a = 5

Therefore, the equation is: x + y = 5

Answer: x + y = 5

Question 13: Find the area of the triangle formed by the lines x = 0, y = 0, and 3x + 4y = 12.

Solution:

The three lines are:

x = 0 (y-axis)

y = 0 (x-axis)

3x + 4y = 12

Finding the vertices of the triangle:

Intersection of x = 0 and y = 0: (0, 0)

Intersection of x = 0 and 3x + 4y = 12: (0, 3)

Intersection of y = 0 and 3x + 4y = 12: (4, 0)

The triangle has vertices at (0, 0), (0, 3), and (4, 0)

Area = (1/2) × base × height = (1/2) × 4 × 3 = 6 square units

Answer: 6 square units

Question 14: Find the equation of the line which cuts off intercepts on the axes whose sum and product are 5 and 6, respectively.

Solution:

Let the x-intercept be a and y-intercept be b

Given: a + b = 5 ... (1)

ab = 6 ... (2)

From equation (1): b = 5 - a

Substituting in equation (2): a(5 - a) = 6

5a - a² = 6

a² - 5a + 6 = 0

(a - 2)(a - 3) = 0

a = 2 or a = 3

If a = 2, then b = 3

If a = 3, then b = 2

Using intercept form x/a + y/b = 1:

Case 1: x/2 + y/3 = 1 → 3x + 2y = 6

Case 2: x/3 + y/2 = 1 → 2x + 3y = 6

Answer: 3x + 2y = 6 or 2x + 3y = 6

Question 15: 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 the x-axis is 30°.

Solution:

The normal form of a line is: x cos α + y sin α = p

Where p is the perpendicular distance from origin and α is the angle the normal makes with positive x-axis

Given: p = 4, α = 30°

x cos 30° + y sin 30° = 4

x(√3/2) + y(1/2) = 4

Multiplying by 2: √3x + y = 8

Answer: √3x + y = 8

Question 16: A line is such that its segment between the axes is bisected at the point (1, 2). Find its equation.

Solution:

Let the line intersect the x-axis at (a, 0) and y-axis at (0, b)

The midpoint of the segment joining (a, 0) and (0, b) is ((a + 0)/2, (0 + b)/2) = (a/2, b/2)

Given that this midpoint is (1, 2):

a/2 = 1 → a = 2

b/2 = 2 → b = 4

Using intercept form: x/a + y/b = 1

x/2 + y/4 = 1

Multiplying by 4: 2x + y = 4

Answer: 2x + y = 4

Question 17: Find the equation of the line passing through (1, 2) and intersecting the x-axis and y-axis at A and B respectively such that the point P(1, 2) divides AB internally in the ratio 2:3.

Solution:

Let A = (a, 0) and B = (0, b)

Using section formula, if P(1, 2) divides AB in ratio 2:3:

P = ((2×0 + 3×a)/(2+3), (2×b + 3×0)/(2+3)) = (3a/5, 2b/5)

Since P = (1, 2):

3a/5 = 1 → a = 5/3

2b/5 = 2 → b = 5

Using intercept form: x/(5/3) + y/5 = 1

3x/5 + y/5 = 1

Multiplying by 5: 3x + y = 5

Answer: 3x + y = 5

Question 18: Find the equation of the line which passes through the point of intersection of 2x + 3y = 5 and 3x - 2y = 1 and is perpendicular to the line x + 2y = 3.

Solution:

First, find the intersection point of 2x + 3y = 5 and 3x - 2y = 1:

From first equation: x = (5 - 3y)/2

Substituting in second equation: 3(5 - 3y)/2 - 2y = 1

(15 - 9y)/2 - 2y = 1

15 - 9y - 4y = 2

15 - 13y = 2

13y = 13

y = 1

x = (5 - 3)/2 = 1

Intersection point = (1, 1)

Slope of line x + 2y = 3 is -1/2

Slope of perpendicular line = -1/(-1/2) = 2

Using point-slope form with point (1, 1) and slope 2:

y - 1 = 2(x - 1)

y - 1 = 2x - 2

y = 2x - 1

Answer: y = 2x - 1 or 2x - y = 1

Question 19: Find the equation of the line passing through (2, 3) and making an angle of 60° with the line x + √3y + 2 = 0.

Solution:

Slope of given line x + √3y + 2 = 0 is -1/√3

Let the slope of required line be m

Using the formula for angle between two lines:

tan 60° = |(m - (-1/√3))/(1 + m(-1/√3))|

√3 = |(m + 1/√3)/(1 - m/√3)|

√3 = |(√3m + 1)/(√3 - m)|

√3(√3 - m) = ±(√3m + 1)

Case 1: 3 - √3m = √3m + 1

3 - 1 = √3m + √3m

2 = 2√3m

m = 1/√3

Case 2: 3 - √3m = -(√3m + 1)

3 - √3m = -√3m - 1

3 = -1 (impossible)

So m = 1/√3

Using point-slope form with point (2, 3) and slope 1/√3:

y - 3 = (1/√3)(x - 2)

√3(y - 3) = x - 2

√3y - 3√3 = x - 2

x - √3y + 3√3 - 2 = 0

Answer: x - √3y + 3√3 - 2 = 0

Question 20: A line is drawn through (1, –2) parallel to the line 2x - 3y + 5 = 0. Find its equation and its distance from the point (4, 1).

Solution:

Slope of line 2x - 3y + 5 = 0 is 2/3

Since parallel lines have same slope, required line has slope 2/3

Using point-slope form with point (1, -2) and slope 2/3:

y - (-2) = (2/3)(x - 1)

y + 2 = (2/3)x - 2/3

y = (2/3)x - 2/3 - 2

y = (2/3)x - 8/3

Multiplying by 3: 3y = 2x - 8

2x - 3y - 8 = 0

Distance from point (4, 1) to line 2x - 3y - 8 = 0:

Distance = |2(4) - 3(1) - 8|/√(2² + (-3)²)

= |8 - 3 - 8|/√(4 + 9)

= |-3|/√13

= 3/√13 = 3√13/13 units

Answer: Equation: 2x - 3y - 8 = 0; Distance: 3√13/13 units