Question

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

Answer 1

STEP 1 : Assumption

Let $A(0,0),B(5,5)$ and $C(-5,5)$ be the given points.

STEP 2 : Finding the distance between the three points

By distance formula distance between two points $P(x_1,y_1)$ and $Q(x_2,y_2)$ is

$PQ=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Therefore distance between $AB,BC$ and $CA$ can be found using the above formula

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

$AB=\sqrt{25+25}$

$\Rightarrow AB=\sqrt{50}$

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

$BC=\sqrt{{10}^2+0}$

$\Rightarrow BC=\sqrt{100}$

$CA=\sqrt{{\left\{0-\left(-5\right)\right\}}^2+{(0-5)}^2}$

$CA=\sqrt{{\left(0+5\right)}^2+{(0-5)}^2}$

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

$CA=\sqrt{25+25}$

$\Rightarrow CA=\sqrt{50}$

Since, $AB=AC$

Hence, $ABC$ is an isosceles triangle.

STEP 3 : Finding the relation between $\mathit{A}\mathit{B}\mathbf{,}\mathbf{}\mathit{B}\mathit{C}$ and $\mathit{C}\mathit{A}$

$A{B}^2=50$

$B{C}^2=100$

$C{A}^2=50$

$\Rightarrow B{C}^2=A{B}^2+C{A}^2$

Hence, $ABC$ is a right angles triangle.

Since, $AB=AC$ and $B{C}^2=A{B}^2+C{A}^2$
Therefore, $ABC$ is a right isosceles triangle.