Question

Find the area of that triangle whose vertices are $(-3,-2),(5,-2)$ and $(5,4)$. Also prove that it is right angle triangle.

Answer 1

$A=\frac{1}{2}\left\{x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right)A=\frac{1}{2}\{-3(-2-4)+5(4+2)+5(-2+2)A=\frac{1}{2}\times 48=24$

$AB=\sqrt{{(-3-5)}^2+{(-2+2)}^2}AB=8BC=\sqrt{{(5-5)}^2+{(-2-4)}^2}BC=6AC=\sqrt{{(-3-5)}^2+{(-2-4)}^2}=10{AC}^2={AB}^2+{BC}^{2}100=64+36100=100$

Triangle satisfies pythagoras theorem

Hence triangle is right angled.