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

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Solution

Use Pythagoras theorem to prove that the points $A (- 2, 3), B (8, 3) and C (6, 7)$ are the vertices of a right triangle.
The distance formula is given by $D = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$ .
Apply the distance formula to find the distance between the vertices $A (- 2, 3) and B (8, 3)$ .
$\begin{array}{rcl} A B & = & \sqrt{{(8 - (- 2))}^{2} + {(3 - 3)}^{2}} \\ \Rightarrow & A B = \sqrt{{(10)}^{2} + {(0)}^{2}} \\ \Rightarrow & A B = 10 \end{array}$
Find the distance between the vertices $B (8, 3) and C (6, 7)$ .
$\begin{array}{rcl} B C & = & \sqrt{{(6 - 8)}^{2} + {(7 - 3)}^{2}} \\ \Rightarrow & B C = \sqrt{{(- 2)}^{2} + {(4)}^{2}} \\ \Rightarrow & B C = \sqrt{20} \end{array}$
Find the distance between the vertices $A (- 2, 3) and C (6, 7)$ .
$\begin{array}{rcl} A C & = & \sqrt{{(6 - (- 2))}^{2} + {(7 - 3)}^{2}} \\ \Rightarrow & A C = \sqrt{{(8)}^{2} + {(4)}^{2}} \\ \Rightarrow & A C = \sqrt{80} \end{array}$
Apply the converse of Pythagoras theorem:
$\begin{array}{rcl} B C^{2} + A C^{2} & = & 80 + 20 \\ \Rightarrow & B C^{2} + A C^{2} = 100 \\ \Rightarrow & B C^{2} + A C^{2} = 10^{2} \\ \Rightarrow & B C^{2} + A C^{2} = A B^{2} \end{array}$
Hence, it is proved that $A (- 2, 3), B (8, 3) and C (6, 7)$ are the vertices of a right triangle.

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