Class 11 RD Sharma Solutions - Chapter 23 The Straight Lines - Exercise 23.9

Chapter 23 of RD Sharma's Class 11 Mathematics textbook covers the concept of straight lines which is a fundamental topic in coordinate geometry. Understanding straight lines is crucial for solving problems related to their equations, slopes, intercepts, and various geometric properties. Exercise 23.9 focuses on applying these concepts to solve practical problems involving straight lines.

The Straight Lines

The Straight lines in coordinate geometry are represented by linear equations and can be analyzed using various properties such as the slope, intercepts, and angle with respect to the axes. Key concepts include:

• Slope: The measure of the steepness of the line, given by the ratio of the change in y to the change in x.
• Intercepts: The points where the line crosses the x-axis and y-axis.
• Equation of a Line: The general form of the line's equation can be written as y=mx+c where m is the slope and c is the y-intercept.

Question 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) slope-intercept form and find slope and y-intercept

Given equation:√3x + y + 2 = 0  

Therefore, y = – √3x – 2

Thus, m = -√3, c = -2

Hence, the slope = – √3 and y-intercept = -2

(ii) Intercept form and find intercept on the axes

Given equation: √3x + y + 2 = 0  

√3x + y = -2  [ Divide both sides by -2 ]

√3x/-2 + y/-2 = 1

Therefore, x-intercept = -2/√3 and y-intercept = -2

(iii) The normal form and find p and α

Given equation: √3x + y + 2 = 0    

-√3x – y = 2  [ Divide both sides by 2 ]

(-√3/2)x - y/2 = 1

The normal form is represented as x cos α + y sin α = p

Thus,

cos α = -√3/2 = cos 210°

sin α = -1/2 = sin 210°

Therefore, p = 1 and α = 210°

Question 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

(iii) x - y + 2√2 = 0

(iv) x - 3 = 0

(v) y - 2 = 0

Solution:

(i) x + √3y – 4 = 0

Given equation: x + √3y – 4 = 0

x + √3y = 4  [ Divide both sides by 2 ]

(1/2)x + (√3/2)y = 2

The normal form is represented as x cos α + y sin α = p

Thus,

cos α = 1/2 = cos 60°

sin α = √3/2 = sin 60°

Therefore, p = 2 and α = 60°

(ii) x + y + √2 = 0

Given equation: x + y + √2 = 0

-x – y = √2  [ Divide both sides by √2 ]

(-1/√2)x - (1/√2)y = 1

The normal form is represented as x cos α + y sin α = p

Thus,

cos α = -1/√2

sin α = -1/√2

Since, both are negative,

Thus α is in III quadrant,

α = π(π/4) = 5π/4 = 225°

Therefore, p = 1 and α = 225°

(iii) x - y + 2√2 = 0

Given equation: x - y + 2√2 = 0

-x + y = 2√2  [ Divide both sides by √2 ]

(-1/√2)x - (-1/√2)y = 2

The normal form is represented as x cos α + y sin α = p

Thus,

cos α = -1/√2 

sin α = -1/√2 

α is in II quadrant,

α = (π/4) + (π/2) = 3π/4 = 135°

Therefore, p = 2 and α = 135°

(iv) x - 3 = 0

Given equation: x - 3 = 0

x = 3

The normal form is represented as x cos α + y sin α = p

Thus,

cos α = 1 = cos 0°

Therefore, p = 3 and α = 0°

(v) y - 2 = 0

Given equation: y - 2 = 0

y = 2

The normal form is represented as x cos α + y sin α = p

Thus,

sin α = 1 = sin π/2°

Therefore, p = 2 and α = π/2°

Question 3: Put the equation x/a + y/b = 1 the slope-intercept form and find its slope and y-intercept?

Solution:

Given equation: x/a + y/b = 1  

Since, the general equation of line is represented as y = mx + c.

Thus,

bx + ay = ab

ay = – bx + ab

y = -bx/a + b

Hence, m = -b/a, c = b

Therefore, the slope = -b/a and y-intercept = b

Question 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 equations: 

3x − 4y + 4 = 0 ...... (i)

2x + 4y − 5 = 0 ...... (ii)

For equation (i),

-3x + 4y = 4 [ Divide both sides by 5 i.e.  √(-3)² + (4)² ]

(-3/5)x + (4/5)y = 4/5

Therefore, p = 4/5

Now for equation (ii),

2x + 4y = – 5

-2x – 4y = 5 [ Divide both sides by √20 i.e.  √(-2)² + (-4)² ]

(-2/√20)x – (4/√20)y = 5/√20

Therefore, p = 5/√20 = 5/4.47

Comparing value of p for equations (i) and (ii) we get,

4/5 < 5/4.47

Therefore, the line 3x − 4y + 4 = 0 is nearest to the origin.

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

Solution:

Given equations: 

4x + 3y + 10 = 0 ...... (i)

5x – 12y + 26 = 0 ...... (ii)

7x + 24y = 50 ...... (iii)

For equation (i),

4x + 3y + 10 = 0

-4x – 3y = 10 [ Divide both sides by 5 i.e.  √(-4)² + (-3)² ]

(-4/5)x – 3/5)y = 10/5

(-4/5)x – 3/5)y = 2

Therefore, p = 2

For equation (ii),

5x − 12y + 26 = 0

-5x + 12y = 26 [ Divide both sides by 13 i.e.  √(-5)² + (12)² ]

(-5/13)x + (12/13)y = 26/13

(-5/13)x + (12/13)y = 2

Therefore, p = 2

For equation (iii), 

7x + 24y = 50 [ Divide both sides by 25 i.e.  √(7)² + (24)² ]

(7/25)x + (24/25)y = 50/25

(7/25)x + (24/25)y = 2

Therefore, p = 2

Hence, the origin is equidistant from the given lines.

Question 6: Find the value of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line √3x + y + 2 = 0?

Solution:

Given equation: √3x + y + 2 = 0

√3x + y = 2

-√3x - y = 2

The normal form is represented as x cos θ + y sin θ = p

Thus,

cos θ = -√3

sin θ = 1 

tan θ = 1/√3

θ = π+(π/6) = 180° + 30° = 210°

Therefore, p = 2 and θ = 120°

Question 7: Reduce the equation 3x - 2y + 6 = 0 to the intercept from and find the x-intercept and y-intercept?

Solution:

Given equation: 3x - 2y + 6 = 0 

-3x + 2y = 6  [ Divide both sides by 6 ]

(-3/6)x + (2/6)y = 6/6

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

Therefore, x-intercept = -2 and y-intercept = 3

Question 8: The perpendicular distance of a line from the origin is 5 units and its slope is -1. Find the equation of the line?

Solution:

Given: Perpendicular distance from the origin to the line is 5 units i.e p = 5

The normal form is represented as x cos α + y sin α = p

x cos α + y sin α = 5

y sin α = -x cos α + 5

y = (-x (cos α/sin α) + 5)

y = -x cot α + 5

Comparing it with the equation y = mx + c,

m = -cot α

-1 = -cot α

cot α = 1

α = π/4

Thus, the equation is,

x cos π/4 + y sin π/4 = 5

x/√2 + y/√2 = 5

x + y = 5√2

Therefore, x + y = 5√2 is the equation of line.

Read More:

• Equation of a Straight Line
• Standard Form of a Straight Line

Conclusion

Exercise 23.9 in Chapter 23 of RD Sharma's textbook helps reinforce the understanding of the straight lines by the applying theoretical concepts to practical problems. The Mastery of these concepts is essential for the solving more complex problems in coordinate geometry and for the preparing for the higher-level mathematics.