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

$$\begin{gathered} \mathrm{The} \mathrm{given} \mathrm{points} \mathrm{are} \mathrm{A}( -2, 3 ) \mathrm{B}( 8, 3 ) \mathrm{and} \mathrm{C}( 6, 7 ). \\ \mathrm{using} \mathrm{distance} \mathrm{firmula}, \mathrm{we} \mathrm{have}: \\ {\mathrm{AB}}^2 =( 8 - ( -2 ){)}^2 +( 3 - 3 {)}^2 \\ \Rightarrow {\mathrm{AB}}^2 = ( 8+ 2 {)}^2 + 0 \\ \Rightarrow {\mathrm{AB}}^2 = {10}^2 \\ \Rightarrow {\mathrm{AB}}^2 = 100 \\ {\mathrm{BC}}^2 = (6-8 {)}^2 + (7-3{)}^2 \\ \Rightarrow {\mathrm{BC}}^2 = ( - 2 {)}^2 + 4^2 \\ \Rightarrow {\mathrm{BC}}^2 = 4+16 \\ \Rightarrow {\mathrm{BC}}^2 = 20 \\ {\mathrm{CA}}^2 = ( -2 -6 {)}^2 + ( 3-7 {)}^2 \\ \Rightarrow {\mathrm{CA}}^2 = (-8{)}^2 +(-4 {)}^2 \\ \Rightarrow {\mathrm{CA}}^2 = 64 + 16 \\ \Rightarrow {\mathrm{CA}}^2 =80 \\ \mathrm{It} \mathrm{can} \mathrm{be} \mathrm{observe} \mathrm{that}: \\ {\mathrm{BC}}^2 + {\mathrm{CA}}^2 = 20 + 80 =100 = {\mathrm{AB}}^2 \\ \mathrm{So}, \mathrm{by} \mathrm{the} \mathrm{converse} \mathrm{of} \mathrm{pythagoras} \mathrm{Theorem}, \\ ∆\mathrm{ABC} \mathrm{is} \mathrm{a} \mathrm{right} \mathrm{angled} \mathrm{triangle} \mathrm{with} \mathrm{right} \mathrm{angled} \mathrm{at} \mathrm{C}. \end{gathered}$$