Answer to Question #350448 in Analytic Geometry for Masukudajijul

Question

Show that the points (-2,0), (2,3) and (5,-1) are the vertices of a right triangle. And find its area

Answer 1

$A(-2,0), B(2,3), C(5, -1)$

$$\overrightarrow{BA}=\langle-2-2, 0-3\rangle=\langle-4, -3\rangle$$

$$\overrightarrow{BC}=\langle5-2, -1-3\rangle=\langle3, -4\rangle$$

$$\overrightarrow{BA}\cdot\overrightarrow{BC}=-4(3)+(-3)(-4)=0$$

Then $\overrightarrow{BA}\perp\overrightarrow{BC},$ and we see that the triangle $ABC$ with vertices $A(-2,0), B(2,3), C(5, -1)$ is a right triangle.

$$|\overrightarrow{BA}|=\sqrt{(-4)^2+(-3)^2}=5$$

$$|\overrightarrow{BC}|=\sqrt{(3)^2+(-4)^2}=5$$

$$Area_{ABC}=\dfrac{1}{2}|\overrightarrow{BA}||\overrightarrow{BA}|$$

$$=\dfrac{1}{2}(5)(5)=12.5({units}^2)$$