## Equation of line Solution11

In this page equation of line solution11 we are going to see solution of each problem with detailed explanation of the worksheet slope of the line.

(4) If the straight lines y/2 = x – p and ax + 5 = 3 y are parallel, then find a

Solution:

y/2 = x – p

y = 2(x-p)

y = 2 x – 2 p

3 y = ax + 5

y = ax/3 + 5/3

m1 = 2

m2 = a/3

If two lines are parallel then

Slope of the first line (m1) = Slope of the second line (m2)

2 = a/3

a = 6

(5) Find the value of a if the straight lines 5 x – 2 y – 9 = 0 and ay + 2 x – 11 = 0 are perpendicular to each other.

Solution:

If two lines are perpendicular then

Slope of the first line (m1) x Slope of the second line (m2) = -1

m1 = -5/(-2)

= 5/2

m2 = -a/2

(5/2) x (-a/2) = -1

-5a/4 = -1

-5a = -4

a=-4/(-5)

a =4/5

(6) Find the value of p for which the straight lines 8 px + (2-3p) y + 1 = 0 and px + 8 y – 7 =0 are perpendicular to each other.

Solution:

Slope of the line 8 px + (2-3p) y + 1 = 0

m1 = -8p/(2-3p)

Slope of the line px + 8 y – 7 =0

m2 = -p/8

If two lines are perpendicular m1 x m2 = -1

-8p/(2-3p) x (-p/8) = -1

p²/(2-3p) = -1

p² = -1(2-3p)

p² = -2 + 3p

p²- 3p + 2 = 0

(p - 1) (p - 2) = 0

p – 1 = 0                p – 2 = 0

p = 1                     p = 2

(7) If the straight line passing through the points (h,3) and (4,1) intersects the line 7 x – 9 y – 19 = 0 t a right angle, find the value of h.

Solution:

The line is passing through the points (h,3) and (4,1) and it is intersecting the line 7 x -9 y – 19 = 0. So we can say that these two lines are perpendicular.

If two lines are perpendicular then

slope of the first line (m₁) x slope of the second line (m₂) = -1

m₁ = (y₂ - y₁)/(x₂ - x₁)

= (1-3)/(4-h)

m₁ = -2/(4-h)

m₂ = -7/(-9)

m₂ = 7/9

[-2/(4-h)]  x [7/9] = -1

[2/(4-h)] x [7/9] = 1

14/9(4-h) = 1

14 = 36 – 9 h

9 h = 36 – 14

9 h = 22

h = 22/9

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