Chapter 1.3, Problem 103E

To determine

To show: the points $A\left(-1,5\right)$ , $B\left(3,7\right)$ and $C\left(5,3\right)$ are the vertices of a right triangle.

Explanation of Solution

Concept Used:

Slope of a line:

The slope of line passing through two point $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by $m=\frac{y_2-y_1}{x_2-x_1}$ .

Two lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

So, to show that the points $A\left(-1,5\right)$ , $B\left(3,7\right)$ and $C\left(5,3\right)$ are the vertices of a right triangle, it is enough to show that the slopes of AB and BC are negative reciprocals of each other.

Now, first find the slope of the line AB through $A\left(-1,5\right)$ and $B\left(3,7\right)$ .

  $\begin{array}{c}{m}_1=\frac{7-5}{3-\left(-1\right)}\\ =\frac{2}{3+1}\\ =\frac{2}{4}\\ =\frac{1}{2}\end{array}$

Now, find the slope of the line BC through $B\left(3,7\right)$ and $C\left(5,3\right)$ .

  $\begin{array}{c}{m}_2=\frac{3-7}{5-3}\\ =\frac{-4}{2}\\ =-2\end{array}$

Clearly $m_1{m}_2=-1$

This implies that the sides AB and BC are perpendicular to each other and hence these are sides of right angled triangle ABC.