Question

Determine whether the points are vertices of a right triangle $A(-3,-4)$, $B(2,6)$ and $C(-6,10)$

Answer 1

Using the distance formula $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$, we get
$A{B}^2=(2+3{)}^2+(6+4{)}^2=5^2+{10}^2=25+100=125$
$B{C}^2=(-6,-2{)}^2+(10-6{)}^2=(-8{)}^2+4^2=64+16=80$
$C{A}^2=(-6+3{)}^2+(10+4{)}^2=(-3{)}^2+(14{)}^2=9+196=205$
i.e., $A{B}^2+B{C}^2=125+80=205=C{A}^2$
Hence ABC is a right angled triangle since the square of one side is equal to sum of the squares of the other two sides.