03Cdistbw2points

= Distance Between Two Points =

Consider the two points A and B on a straight line (shown right).
 * A (x 1, y 1 )
 * B (x 2, y 2 )

If we construct a right angled triangle, ABC then the point C will be at (x 2, y 1 )

The lengths of the two sides will therefore be:
 * AC = x 2 – x 1
 * CB = y 2 – y 1

We can therefore find the length of AB using Pythagoras' Theorem:

math \\ . \qquad c^2 = a^2 + b^2 \\. \\ . \qquad AB^2 = AC^c + CB^2 \\. \\ . \qquad AB^2 = \big( x_2 - x_1 \big)^2 + \big( y_2 - y_1 \big)^2 math

Hence math . \qquad AB = \sqrt{ \big( x_2 - x_1 \big)^2 + \big( y_2 - y_1 \big)^2 } math

This formula can be used to find the length of a straight line between any two points.

** Example 1 **
Find the distance between A (–2, 6) and B (3, 2)
 * Solution: **

math \\ . \qquad A = (-2, \; 6) = (x_1, \; y_1 ) \\. \\ . \qquad B = (3, \; 2) = (x_2, \; y_2 ) math

math \\ . \qquad AB = \sqrt{ \big( x_2 - x_1 \big)^2 + \big( y_2 - y_1 \big)^2 } \\. \\ . \qquad AB = \sqrt{ \big( 3 - (-2) \big)^2 + \big( 2 - 6 \big)^2 } \\. \\ . \qquad AB = \sqrt{ \big( 5 \big)^2 + \big( -4 \big)^2 } \\. \\ . \qquad AB = \sqrt{ 25 + 16 } \\. \\ . \qquad AB = \sqrt{41} math

** Example 2 **
What type of triangle is formed by the points A (1, 1), B (3, –1) and C (–1, –3)


 * Solution: **

The triangle could be:
 * Equilateral (3 sides equal in length)
 * Isosceles (2 sides equal in length)
 * Scalene (no sides equal in length)

Length AB math \\ . \qquad AB = \sqrt{ \big( 3-1 \big)^2 + \big( (-1)-1 \big)^2 } \\. \\ . \qquad \quad = \sqrt{ \big( 2 \big)^2 + \big( -2 \big)^2 } \\. \\ . \qquad \quad = \sqrt{ 4 + 4 } \\. \\ . \qquad \quad = \sqrt{ 8 } \\. \\ . \qquad \quad = 2 \sqrt{ 2 } math

Length AC math \\ . \qquad AC = \sqrt{ \big( (-1)-1 \big)^2 + \big( (-3)-1 \big)^2 } \\. \\ . \qquad \quad = \sqrt{ \big( -2 \big)^2 + \big( -4 \big)^2 } \\. \\ . \qquad \quad = \sqrt{ 4 + 16 } \\. \\ . \qquad \quad = \sqrt{ 20 } \\. \\ . \qquad \quad = 2 \sqrt{ 5 } math

Length BC math \\ . \qquad BC = \sqrt{ \big( (-1)-3 \big)^2 + \big( (-3)-(-1) \big)^2 } \\. \\ . \qquad \quad = \sqrt{ \big( -4 \big)^2 + \big( -2 \big)^2 } \\. \\ . \qquad \quad = \sqrt{ 16 + 4 } \\. \\ . \qquad \quad = \sqrt{ 20 } \\. \\ . \qquad \quad = 2 \sqrt{ 5 } math

Clearly AC = BC but AB is different

Hence triangle ABC is isosceles. .