Home > Graphing > Distance between two points

## Distance between two points

Let’s start by considering the points $(x_1,y_1)$ and $(x_2,y_2)$

We have used the two points to construct a right-angled triangle.

The horizontal distance is equal to $x_2-x_1$

The vertical distance is equal to $y_2-y_1$

So using Pythagoras’ theorem we can say the distance between the points $=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

If we define the change in x as $\delta x$ and the change in y as $\delta y$ the the distance become s$=\sqrt{(\delta x)^2+(\delta y)^2}$

Examples

Find the distance between the two points $(3,7)$ and $(7,-2)$

$\delta x = 3-7=-4$.  Since this is the length of a side of a triangle we will positive 4.

$\delta y = 7--2=9$

$\therefore$ distance $= \sqrt{4^2+7^2}=\sqrt{65}$

A triangle is made with the three points $A=(2,8),B=(5,12)$ and $C=(6,11)$.  Show that the triangle is an isosceles triangle.

An isosceles triangle has two sides of equal length, if we can show that two of our sides are equal then we have proved it is isosceles.

The length AB =$\sqrt{3^2+4^2}=5$

The length BC =$\sqrt{1^2+1^2}=\sqrt{2}$

The length AC=$\sqrt{4^2+3^2}=5$

Since the length AB = the length AC, the triangle is isosceles.

Advertisements
Categories: Graphing
1. No comments yet.
1. No trackbacks yet.