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Examples of Two Point Formula

We earlier saw the explanation of two point formula for calculating the equation of a straight line. Here are a few more examples of two point formula.

In the diagram below, the points A (-2, 0), B (1, 3) and C (5, 0) are the coordinates of a triangle ABC.

example_of_two_point_formula

 

a. Find the gradient of each side of the triangle ABC.

Gradient of AB = mAB = {y_2~-~y_1}/{x_2~-~x_1}

=  {3~-~0}/{1~-~(-1)}

=  3/3  =  1

Gradient of BC = mBC = {y_2~-~y_1}/{x_2~-~x_1}

=  {0~-~3}/{5~-~1}

=  –3/4

Gradient of AC = mAC = {0~-~0}/{5~-~(-2)}

=  0

 

b. Find the equation of each lines AB, BC and CA.

Equation of line AB

mAB = 1 and passes through A (-2, 0)

Using the point gradient formula

(y – 0) = 1(x – (-2))

y = x + 2

Equation of line BC

mBC = –3/4 and passes through B (1, 3)

Using the point gradient formula

(y – 3) = –3/4 (x – 1)

4(y – 3) = -3(x – 1)

4y – 12 = -3x + 3

3x + 4y – 12 – 3 = 0

3x + 4y – 15 = 0

Equation of line CA

mCA = 0 and passes through A (-2, 0)

(y – 0) = 0

y = 0 (or x-axis)

 

c. What is the y-intercept of each of the lines AB, BC and CA.

Equation of line AB is y = x + 2.

Therefore the y-intercept of AB = 2

Equation of line BC is 3x + 4y – 15 = 0

We find the y-intercept of BC when x = 0

4y – 15 = 0

Or y = 15/4 – y-intercept

Equation of line CA is y = 0

So y-intercept = 0

 

d. Find the equation of the line passing through the midpoint of AB and C.

A: (-2, 0) and B: (1, 3)

Midpoint of AB = ({-2~+~1}/2,~{0~+~3}/2)  =  ({-1}/2,~3/2)

Let’s call the above midpoint D

The line joining D ({-1}/2,~3/2) and C (5, 0) is

y – 0 =  {0~-~3/2}/{5~-~-1/2} (x – 5)

y =  -{3/2}/{11/2} (x – 5)

=  -{3/11} (x – 5)

11y = -3x + 15 or

3x + 11y = 15