Question

Show that the points $(-2,3),(8,3)$ and $(6,7)$ are the vertices of a right triangle.

Answer 1

Given coordinetes of triangle $A(-2,3),B(8,3)$ and $C(6,7)$

$AB=\sqrt{{[8-(-2\left)\right]}^2+(3-3{)}^2}=\sqrt{(10{)}^2+(0{)}^2}=\sqrt{100}$

$AC=\sqrt{{[6-(-2\left)\right]}^2+(7-3{)}^2}=\sqrt{(8{)}^2+(4{)}^2}=\sqrt{64+16}=\sqrt{80}$

$BC={\sqrt{(6-8)+[7-(2\left)\right]}}^2=\sqrt{(-2{)}^2+(4{)}^2}=\sqrt{4+16}=\sqrt{20}$

If $ABC$ is a right angled triangle, then square of one side must be equal to sum of suare of other two sides.

$A{B}^2=(\sqrt{100}{)}^2=100$

and $A{C}^2+B{C}^2=(\sqrt{80}{)}^2+(\sqrt{20}{)}^2=100$

$\therefore A{B}^2=A{C}^2+B{C}^2$
thus triangle satisfies the Pythagoras theorem

Hence, the triangle is right angled triangle.