Question

Prove that the points (7,10), (-2,5) and (3,-4) are the vertices of an isosceles right triangle.

Answer 1

Let the given points be A (7,10) B(-2,5) and C(3,-4).

Using distance formula, we have

$AB=\sqrt{(7+2{)}^2+(10-5{)}^2}=\sqrt{81+25}=\sqrt{106}BC=\sqrt{(-2-3{)}^2+(5+4{)}^2}=\sqrt{25+81}=\sqrt{106}CA=\sqrt{(7-3{)}^2+(10+4{)}^2}=\sqrt{16+196}=\sqrt{212}$
Since AB = BC, therefore, $\Delta ABC$ is an isosceles triangle.
Also, $A{B}^2+B{C}^2=106+106=212=A{C}^2$
So, $\Delta ABC$ is a right triangle right angled at $\angle B.$
So, $\Delta ABC$ is an isosceles triangle as well as well as a right triangle.
Thus, the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right triangle.