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Examples of Equations of Perpendicular Lines

We learnt from the equations of perpendicular lines that given two lines with gradients m1 and m2 respectively, the two lines will be perpendicular when the product of their gradients is equal to -1. That is –

m1m =  -1  or  m1  =  {-1}/{m_2}

We now look at a few examples of equations of perpendicular lines, and how to solve them.

 

Example 1: Find the equation of a line that is perpendicular to 2x – 5y + 7 = 0 and passing through the point (2, 3)

We can solve this by two methods.

Method 1: First find the gradient of the given line.

2x – 5y + 7 = 0

2x + 7 = 5y

y = {2/5}x~+~{7/5}

So m1 =  2/5

Next find the gradient of the line perpendicular to 2x – 5y + 7 = 0.

m2  =  {-1}/{m_1}

=  {-1}/{2/5}

=  {-5}/2

Given a point (2, 3) and gradient m2  =  {-5}/2, use the point gradient form to find the equation of the straight line.

Equation of the required line is –

y – y1  =  m(x – x1)

y – 3  =  {-5}/2 (x – 2)

Multiplying by 2 on both sides –

2y – 6  =  -5(x – 2)

2y – 6  =  -5x + 10

5x + 2y – 6 – 10 = 0

5x + 2y – 16 = 0 is the equation of the line perpendicular to 2x – 5y + 7 = 0 and passing through the point (2, 3).

 

Method 2: Using perpendicular property of lines, the line perpendicular to 2x – 5y + 7 = 0 is 5x + 2y + k = 0  (interchange of the x and y coefficients and change only one of the coefficients to the opposite sign).

Substitute (2, 3) in this equation and solve for k. So

5x + 2y + k = 0

5 x 2 + 2 x 3 + k = 0

10 + 6 + k = 0

k = -16

So the required equation of the line perpendicular to 2x – 5y + 7 = 0 and passing through (2, 3) is 5x + 2y – 16 = 0

 

 Example 2: Show that the line ax + by + c = 0 is

  1. parallel to ax + by + d = 0, and
  2. perpendicular to bx – ay + e = 0

Given Line 1 to be ax + by + c = 0.

Gradient of Line 1 is

by = – c – ax

y = {-c/b}~-~{-a/b}x

m1 =  {-a/b}         Equation 1

Gradient of Line 2 is

by = – d – ax

y = {-d/b}~-~{-a/b}x

m2 =  {-a/b}         Equation 2

Gradient of Line 3 is –

ay =  bx + e

y = {b/a}x~-~{e/a}

m3 =  b/a          Equation 3

From 1 and 2, m1 = m2 =  {-a/b}

Hence lines 1 and 2 are parallel.

From 1 and 3, m1 =  {-a/b}  =  {-1}/{b/a}

=  {-1}/{m_3}

m1 =  {-1}/{m_3}

Or m1 x m3 = -1.

Hence lines 1 and 3 are perpendicular to each other.

 

Example 3: A line is parallel to 4x – 5y + 1 = 0 and passes through the points (1, 3) and (5, a). Find the value of ‘a’.

Gradient of the line 4x – 5y + 1 = 0 is

5y = 4x + 1

y =  {4/5}x~+~{1/5}

m1 =  4/5.

The line parallel to 4x – 5y +1 = 0 also has a gradient of 4/5.

Since the line passes through (1, 3) and (5, a), the gradient of the line passing through these points is also 4/5. So

gradient = m = {y_2~-~y_1}/{x_2~-~x_1}

4/5  =  {a~-~3}/{5~-~1}

{a~-~3}/4  =  4/5

5(a – 3) = 4 x 4

5a – 15  =  16

5a  =  31

a = 31/5

So the value of a = 31/5