Prove that the points (0, 0), (5, 5) and (−5, 5)
Question:

Prove that the points (0, 0), (5, 5) and (−5, 5) are the vertices of a right isosceles triangle.

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

In an isosceles triangle there are two sides which are equal in length.

By Pythagoras Theorem in a right-angled triangle the square of the longest side will be equal to the sum of squares of the other two sides.

Here the three points are A(0,0), B(5,5) and C(5,5).

Let us check the length of the three sides of the triangle.

$A B=\sqrt{(0-5)^{2}+(0-5)^{2}}$

$=\sqrt{(-5)^{2}+(-5)^{2}}$

$=\sqrt{25+25}$

$A B=5 \sqrt{2}$

$B C=\sqrt{(5+5)^{2}+(5-5)^{2}}$

$=\sqrt{(10)^{2}+(0)^{2}}$

$=\sqrt{25+25}$

$A C=5 \sqrt{2}$

Here, we see that two sides of the triangle are equal. So the triangle formed should be an isosceles triangle.

Further it is seen that

If in a triangle the square of the longest side is equal to the sum of squares of the other two sides then the triangle has to be a right-angled triangle.

Hence proved that the triangle formed by the three given points is an.