Question

Check whether $(5,-2),(6,4)$ and $(7,-2)$ are the vertices of an isosceles triangle.

Answer 1

The vertices of a triangle ABC is given as $A\left(5,-2\right)$, $B\left(6,4\right)$ and $C\left(7,-2\right)$.

The distance AB is,

$AB=\sqrt{{\left(6-5\right)}^2+{\left(4-\left(-2\right)\right)}^2}$

$=\sqrt{{\left(1\right)}^2+{\left(6\right)}^2}$

$=\sqrt{1+36}$

$=\sqrt{37}$

The distance BC is,

$BC=\sqrt{{\left(7-6\right)}^2+{\left(-2-4\right)}^2}$

$=\sqrt{{\left(1\right)}^2+{\left(-6\right)}^2}$

$=\sqrt{1+36}$

$=\sqrt{37}$

The distance CA is,

$CA=\sqrt{{\left(5-7\right)}^2+{\left(-2-\left(-2\right)\right)}^2}$

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

$=\sqrt{4}$

$=2$

Since, $AB=BC$, that is the two sides of a triangle are equal, therefore, the triangle is an isosceles triangle.