Answer to Question #295684 in Trigonometry for Manot

Question

Show that the points (-3,9), (3,5) and (1,2), are the vertices of a right triangle and find its area.

Answer 1

Let us consider that the given points are : $A = (x_1, y_1) = (3,5) ,B = (x_2, y_2) = (-3,9) \space and \space C = (x_3,y_3) = (1,2)$

The same has been shown in the figure below.

So, we get :

$AB^2 = (y_2-y_1)^2 + (x_2-x_1)^2 \newline= (9-5)^2 + (-3-3)^2 = 4^2 + -6^2 = 16 + 36 = 52$

$BC^2 = (y_3-y_2)^2 + (x_3-x_2)^2 \newline(2-9)^2 + (1+3)^2 = -7^2 + 4^2 = 49 + 16 = 65$

$AC^2 = (y_3-y_1)^2 + (x_3-x_1)^2 \newline (2-5)^2 + (1-3)^2 = -3^2 + -2^2 = 9 + 4 = 13$

Where AB, BC and AC is the length of the sides of triangle ABC.

Now what we see from the values of AB, BC and AC is :

$65 = 52 + 13 \implies BC^2 = AB^2 + AC^2$

This means ABC is a right angled triangle having hypotenuse BC, base AC and perpendicular as AB. So we get $\angle BAC = 90 \degree$. This has been shown in figure also. And it means that the given points A, B, C are the vertices of a right angle triangle.

Now the area of right angel triangle is :

$Area = (1/2)\cdot base\cdot height = (1/2)\cdot (AC)\cdot (AB) \newline= (1/2)\cdot \sqrt{13}\cdot \sqrt{52} \newline= (1/2)\cdot\sqrt{(13\cdot52)} \newline= (1/2)\cdot\sqrt{(13\cdot13\cdot4)} \newline= (1/2)\cdot(\sqrt{(13^2\cdot2^2) }= (1/2)\cdot13\cdot2 = 13\space square\space units$

So the area of right angle triangle ABC is 13 square units.