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

• Last Updated : 25 Mar, 2022

### Question 1: Reduce the equation âˆš3x + 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 Î±.

Solution:

(i) slope-intercept form and find slope and y-intercept

Given equation:âˆš3x + y + 2 = 0

Therefore, y = â€“ âˆš3x â€“ 2

Thus, m = -âˆš3, c = -2

Hence, the slope = â€“ âˆš3 and y-intercept = -2

(ii) Intercept form and find intercept on the axes

Given equation: âˆš3x + y + 2 = 0

âˆš3x + y = -2  [ Divide both sides by -2 ]

âˆš3x/-2 + y/-2 = 1

Therefore, x-intercept = -2/âˆš3 and y-intercept = -2

(iii) The normal form and find p and Î±

Given equation: âˆš3x + y + 2 = 0

-âˆš3x â€“ y = 2  [ Divide both sides by 2 ]

(-âˆš3/2)x – y/2 = 1

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

Thus,

cos Î± = -âˆš3/2 = cos 210Â°

sin Î± = -1/2 = sin 210Â°

Therefore, p = 1 and Î± = 210Â°

### Question 2: Reduce the following equations to the normal form and find p and Î± in each case:

(i) x + âˆš3y â€“ 4 = 0

(ii) x + y + âˆš2 = 0

(iii) x – y + 2âˆš2 = 0

(iv) x – 3 = 0

(v) y – 2 = 0

Solution:

(i) x + âˆš3y â€“ 4 = 0

Given equation: x + âˆš3y â€“ 4 = 0

x + âˆš3y = 4  [ Divide both sides by 2 ]

(1/2)x + (âˆš3/2)y = 2

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

Thus,

cos Î± = 1/2 = cos 60Â°

sin Î± = âˆš3/2 = sin 60Â°

Therefore, p = 2 and Î± = 60Â°

(ii) x + y + âˆš2 = 0

Given equation: x + y + âˆš2 = 0

-x â€“ y = âˆš2  [ Divide both sides by âˆš2 ]

(-1/âˆš2)x – (1/âˆš2)y = 1

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

Thus,

cos Î± = -1/âˆš2

sin Î± = -1/âˆš2

Since, both are negative,

Thus Î± is in III quadrant,

Î± = Ï€(Ï€/4) = 5Ï€/4 = 225Â°

Therefore, p = 1 and Î± = 225Â°

(iii) x – y + 2âˆš2 = 0

Given equation: x – y + 2âˆš2 = 0

-x + y = 2âˆš2  [ Divide both sides by âˆš2 ]

(-1/âˆš2)x – (-1/âˆš2)y = 2

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

Thus,

cos Î± = -1/âˆš2

sin Î± = -1/âˆš2

Î± = (Ï€/4) + (Ï€/2) = 3Ï€/4 = 135Â°

Therefore, p = 2 and Î± = 135Â°

(iv) x – 3 = 0

Given equation: x – 3 = 0

x = 3

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

Thus,

cos Î± = 1 = cos 0Â°

Therefore, p = 3 and Î± = 0Â°

(v) y – 2 = 0

Given equation: y – 2 = 0

y = 2

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

Thus,

sin Î± = 1 = sin Ï€/2Â°

Therefore, p = 2 and Î± = Ï€/2Â°

### 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 = â€“ 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 â€“ 4y + 4 = 0 and 2x + 4y â€“ 5 = 0 to the normal form and hence find which line is nearer to the origin?

Solution:

Given equations:

3x âˆ’ 4y + 4 = 0 …… (i)

2x + 4y âˆ’ 5 = 0 …… (ii)

For equation (i),

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

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

Therefore, p = 4/5

Now for equation (ii),

2x + 4y = â€“ 5

-2x â€“ 4y = 5 [ Divide both sides by âˆš20 i.e.  âˆš(-2)2 + (-4)2 ]

(-2/âˆš20)x â€“ (4/âˆš20)y = 5/âˆš20

Therefore, p = 5/âˆš20 = 5/4.47

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

4/5 < 5/4.47

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

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

Solution:

Given equations:

4x + 3y + 10 = 0 …… (i)

5x â€“ 12y + 26 = 0 …… (ii)

7x + 24y = 50 …… (iii)

For equation (i),

4x + 3y + 10 = 0

-4x â€“ 3y = 10 [ Divide both sides by 5 i.e.  âˆš(-4)2 + (-3)2 ]

(-4/5)x â€“ 3/5)y = 10/5

(-4/5)x â€“ 3/5)y = 2

Therefore, p = 2

For equation (ii),

5x âˆ’ 12y + 26 = 0

-5x + 12y = 26 [ Divide both sides by 13 i.e.  âˆš(-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.  âˆš(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 Î¸ and p, if the equation x cos Î¸ + y sin Î¸ = p is the normal form of the line âˆš3x + y + 2 = 0?

Solution:

Given equation: âˆš3x + y + 2 = 0

âˆš3x + y = 2

-âˆš3x – y = 2

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

Thus,

cos Î¸ = -âˆš3

sin Î¸ = 1

tan Î¸ = 1/âˆš3

Î¸ = Ï€+(Ï€/6) = 180Â° + 30Â° = 210Â°

Therefore, p = 2 and Î¸ = 120Â°

### 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 Î± + y sin Î± = p

x cos Î± + y sin Î± = 5

y sin Î± = -x cos Î± + 5

y = (-x (cos Î±/sin Î±) + 5)

y = -x cot Î± + 5

Comparing it with the equation y = mx + c,

m = -cot Î±

-1 = -cot Î±

cot Î± = 1

Î± = Ï€/4

Thus, the equation is,

x cos Ï€/4 + y sin Ï€/4 = 5

x/âˆš2 + y/âˆš2 = 5

x + y = 5âˆš2

Therefore, x + y = 5âˆš2 is the equation of line.

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