FINDING EQUATION OF A PERPENDICULAR LINE

Write down the equation of each of the following lines.

Example 1 :

Perpendicular to y = 2x + 4 and passing through (0, 3)

Solution :

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

Comparing slope intercept form y = mx + c in (1), we get

Slope (m₁) = 2

Since the two lines are perpendicular, then the product of their slopes is equal to - 1.

m₁ x m₂  =  -1 

2 x m₂  =  -1

m₂  = -1/2

Using Point Slope Form :

y - y₁ = m(x - x₁)

We have,

Slope (m) = -1/2, Point (x₁, y₁) = (0, 3)

y - 3 = -1/2(x - 0)

y - 3 = -1/2x

2(y - 3) = -x

2y - 6 = -x

x + 2y - 6 = 0

So, the equation of the line is x + 2y - 6 = 0.

Example 2 :

Perpendicular to y = -3x - 8 and passing through (0, -2)

Solution :

y = -3x - 8 ----(1)

Comparing slope intercept form y = mx + c in (1), we get

Slope (m₁) = -3

Since the two lines are perpendicular, then the product of their slopes is equal to - 1.

m₁ x m₂  =  -1 

-3 x m₂  =  -1

m₂  = 1/3

Using Point Slope Form :

y - y₁ = m(x - x₁)

We have,

Slope (m) = 1/3, Point (x₁, y₁) = (0, -2)

y + 2 = 1/3(x - 0)

y + 2 = 1/3x

3(y + 2) = x

3y + 6 = x

x - 3y - 6 = 0

So, the equation of the line is x - 3y - 6 = 0.

Example 3 :

Perpendicular to x + y = 6 and passing through (0, 1)

Solution :

x + y = 6

y = -x + 6 ----(1)

Comparing slope intercept form y = mx + c in (1), we get

Slope (m₁) = -1

m₁ x m₂  =  -1

-1 x m₂  =  -1

m₂  = 1

Using Point Slope Form :

y - y₁ = m(x - x₁)

We have,

Slope (m) = 1, Point (x₁, y₁) = (0, 1)

y - 1 = 1(x - 0)

y - 1 = x

x - y + 1 = 0

So, the equation of the line is x - y + 1 = 0.

Example 4 :

Perpendicular to y = (1/3)x - 2 and passing through the origin

Solution :

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

Comparing slope intercept form y = mx + c in (1), we get

Slope (m₁) = 1/3

m₁ x m₂  =  -1

1/3 x m₂  =  -1

m₂  = -3

Using Point Slope Form :

y - y₁ = m(x - x₁)

We have,

Slope (m) = -3, Point (x₁, y₁) = (0, 0)

y - 0 = -3(x - 0)

y - 0 = -3x

3x + y - 0 = 0

So, the equation of the line is 3x + y - 0 = 0.

Example 5 :

Perpendicular to y = (-1/5)x + 8 and passing through (0, -2)

Solution :

y = (-1/5)x + 8 ----(1)

Comparing slope intercept form y = mx + c in (1), we get

Slope (m₁) = -1/5

m₁ x m₂  =  -1

-1/5 x m₂  =  -1

m₂  = 5

Using Point Slope Form :

y - y₁ = m(x - x₁)

We have,

Slope (m) = 5, Point (x₁, y₁) = (0, -2)

y + 2 = 5(x - 0)

y + 2 = 5x

5x - y - 2 = 0

So, the equation of the line is 5x - y - 2 = 0.

Example 6 :

Perpendicular to y = (-2/9)x - 10 and passing through (0, 6)

Solution :

y = (-2/9)x - 10 ----(1)

Comparing slope intercept form y = mx + c with (1), we get

Slope (m₁) = -2/9

m₁ x m₂  =  -1

-2/9 x m₂  =  -1

m₂  = 9/2

Using Point Slope Form :

y - y₁ = m(x - x₁)

We have,

Slope (m) = 9/2, Point (x₁, y₁) = (0, 6)

y - 6 = (9/2)(x - 0)

y - 6 = (9/2)x

2(y - 6) = 9x

2y - 12 = 9x

9x - 2y + 12 = 0

So, the equation of the line is 9x - 2y + 12 = 0.