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Distance between two points

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

Coordinate Pythagoras

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}


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.

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