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

**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-interceptGiven 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 axesGiven 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 αGiven 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 = 0Given 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 = 0Given 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 ar 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 = 0Given 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 = 0Given 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 = 0Given 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 gey,

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.