Question

Prove that A(2, 1), B(0, 3) and C(–2, 1) are the vertices of an isosceles right angled triangle.

Answer 1

Answer:

You need to find distance between point A and B, then point B and C, then between point C and A using distance formula i.e.,$\sqrt{(x_{2}-x_{1} )^{2} +(y_{2}-y_{1}) ^{2} }$

Step-by-step explanation:

Answer 2

A(2, 1), B(0, 3) and C(–2, 1) are the vertices of an isosceles right angled triangle, proved.

Step-by-step explanation:

Given,

The vertices of an isosceles right angled triangle are A(2, 1), B(0, 3) and C(– 2, 1).

To prove that,  A(2, 1), B(0, 3) and C(–2, 1) are the vertices of an isosceles right angled triangle.

Using the distance formula,

$\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

AB = $\sqrt{(0-2)^2+(3-1)^2}$

= $\sqrt{(-2)^2+(2)^2}=\sqrt{4+4} =2\sqrt{2}$

BC = $\sqrt{(-2-0)^2+(1-3)^2}$

= $\sqrt{(-2)^2+(-2)^2}=\sqrt{4+4} =2\sqrt{2}$

Also,

BC = $\sqrt{(2+2)^2+(1-1)^2}$

$=\sqrt{(4)^2+(0)^2}=\sqrt{16+0} =\sqrt{16} =4$

∴ AB = BC $=2\sqrt{2}$, two sides are equal, proved.

Thus,  A(2, 1), B(0, 3) and C(–2, 1) are the vertices of an isosceles right angled triangle, proved.