Question

Prove that the points $(0,7,-10),(1,6,-6)$ and $(4,9,-6)$ are the vertices of an isosceles triangle.

Answer 1

Let $A=(0,7,-10)$, $B=(1,6,-6)$ AND $C=(4,9,-6)$ then:

$AB=\sqrt{(-1{)}^2+(1{)}^2+(-4{)}^2}=\sqrt{18}$

$BC=\sqrt{(-3{)}^2+(-3{)}^2+(0{)}^2}=\sqrt{18}$

$CA=\sqrt{(-4{)}^2+(-2{)}^2+(-4{)}^2}=6$

Since these lengths satisfy the triangle inequality and $AB=BC$, it is clear that these points form an isosceles triangle.