Chapter 4.6, Problem 13PSB

To determine

To Prove: $\Delta ABC$ is right angle.

Explanation of Solution

Given information: coordinates of A, B and C are given.

Formula used: Slope of line joining $(x_1,y_1)$ and $(x_2,y_2)$ is $\frac{(y_2-y_1)}{(x_2-x_1)}$ . If product of slopes of two lines is -1, then the lines are perpendicular.

Calculation:

  $\begin{array}{l}\text{Slope}\text{of}AB=\frac{5-11}{1-6}=\frac{6}{5}.\\ \text{Slope}\text{of}BC=\frac{0-5}{7-1}=\frac{-5}{6}.\\ \text{product}\text{of}\text{slopes}\text{of}\text{AB}\text{and}\text{BC}=\frac{-5}{6}\times \frac{6}{5}=-1\\ \therefore AB\perp BC\\ \text{Hence}\angle ABC\text{is}\text{right}\text{triangle}\text{with}\text{hypotenuse}\text{AC}\text{.}\end{array}$