Question

Prove that the points $A(1,-3),B(-3,0)$ and $C(4,1)$ are the vertices of right isosceles triangle.

Answer 1

We know that the distance between the two points $(x_1,y_1)$ and $(x_2,y_2)$ is

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

Let the given vertices be $A=(1,-3)$, $B=(-3,0)$ and $C=(4,1)$

We first find the distance between $A=(1,-3)$ and $B=(-3,0)$ as follows:

$AB=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(-3-1)}^2+{(0-(-3\left)\right)}^2}=\sqrt{{(-4)}^2+{(0+3)}^2}=\sqrt{{(-4)}^2+3^2}=\sqrt{16+9}$

$=\sqrt{25}=5$

Similarly, the distance between $B=(-3,0)$ and $C=(4,1)$ is:

$BC=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(4-(-3\left)\right)}^2+{(1-0)}^2}=\sqrt{{(4+3)}^2+1^2}=\sqrt{7^2+1^2}=\sqrt{49+1}=\sqrt{50}$

$=\sqrt{5^2\times 2}=5\sqrt{2}$

Now, the distance between $C=(4,1)$ and $A=(1,-3)$ is:

$CA=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(1-4)}^2+{(-3-1)}^2}=\sqrt{{(-3)}^2+{(-4)}^2}=\sqrt{9+16}=\sqrt{25}=\sqrt{5^2}=5$

We also know that If any two sides have equal side lengths, then the triangle is isosceles.

Here, since the lengths of the two sides are equal that is $AB=CA=5$, therefore, $ABC$ is an isosceles triangle.

Now consider,

$A{B}^2+C{A}^2=5^2+5^2=25+25=50=(5\sqrt{2}{)}^2=B{C}^2$

Since $A{B}^2+C{A}^2=B{C}^2$, therefore, $ABC$ is a right angled triangle.

Hence, the given vertices are the vertices of isosceles right angled triangle.