Question

Prove that $(2,-2),(-2,1)$ and $(5,2)$ are the vertices of a right angled-triangle. Find the area of the triangle and the length of the hypotenuse.

Answer 1

To prove:

$A(2,-2),B(-2,1)$ and C(5,2) are the vertices of a right angled-triangle.

Proof:

By distance formula,

$AB=\sqrt{(-2-2{)}^2+(1+2{)}^2}=\sqrt{16+9}=\sqrt{25}=5$

$BC=\sqrt{(-2-5{)}^2+(1-2{)}^2}=\sqrt{(-7{)}^2+(-1{)}^2}=\sqrt{50}=5\sqrt{2}$ Length of the hypotenuse

$AC=\sqrt{(5-2{)}^2+(2+2{)}^2}=\sqrt{(3{)}^2+(4{)}^2}=\sqrt{25}$

$\therefore A{B}^2+A{C}^2=25+25=50=B{C}^2$

$\therefore$ $△$ ABC is a right-angled triangle

Area of the triangle ABC

= $\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$sq.

Area of the $△$ABC

$=\frac{1}{2}\times Base\times height$

$=\frac{1}{2}\times 5\times 5$

$=\frac{25}{2}$sq.units.