Question

Show that the points $(12,8),(-2,6),(6,0)$ are vertices of a right triangle.

Answer 1

Step-1: Finding the length of the sides.

Let three points be $A(12,8),B(-2,6),C(6,0)$

We know the distance between two points $\left(X_1,Y_1\right)and\left(X_2,Y_2\right)$ is given by $S=\sqrt{{(X_2-X_1)}^2+{(Y_2-Y_1)}^2}$

Distance $$\begin{gathered} AB=\sqrt{{\{12-(-2\left)\right\}}^2+(8-6{)}^2} \\ AB=\sqrt{200} \end{gathered}$$

Distance $$\begin{gathered} AC=\sqrt{{(12-6)}^2+{(8-0)}^2} \\ AC=10 \end{gathered}$$

Distance $$\begin{gathered} BC=\sqrt{{\left\{\right(-2)-6\}}^2+(6-0{)}^2} \\ BC=10 \end{gathered}$$

Step-2: Use Pythagoras' theorem

In a right angle triangle ${\left(hypotenuse\right)}^2={\left(perpendicular\right)}^2+(base{)}^2$

AB being the largest side can be considered as the hypotenuse.

So, here $A{B}^2=200$

$$\begin{gathered} A{C}^2+B{C}^2=100+100=200 \\ \Rightarrow A{B}^2=A{C}^2+B{C}^2 \end{gathered}$$

As the triangle formula by the given points obeys the Pythagoras theorem.

Hence, the given points $(12,8),(-2,6),(6,0)$ are vertices of a right-angle triangle.