Chapter 13, Problem 19CT

To determine

To show: The triangle JKL is a right-angled triangle with the help of slopes.

Explanation of Solution

Given information: The points are $J\left(-12,0\right)$ , $K\left(0,6\right)$ and $L\left(-3,-3\right)$ .

Formula used: The slope m of a line between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ can be calculated by the formula:

  $m=\frac{y_2-y_1}{x_2-x_1}$

Proof:

The sides of $\Delta JKL$ are JK , KL and LJ .

To find the slope of line JK , substitute 6 for $y_2$ , 0 for $y_1$ , 0 for $x_2$ and $-12$ for $x_1$ in the formula.

  $\begin{array}{c}{m}_1=\frac{6-0}{0-\left(-12\right)}\\ =\frac{6}{12}\\ =\frac{1}{2}\end{array}$

To find the slope of line KL , substitute $-3$ for $y_2$ , 6 for $y_1$ , $-3$ for $x_2$ and 0 for $x_1$ in the formula.

  $\begin{array}{c}{m}_2=\frac{\left(-3\right)-6}{\left(-3\right)-0}\\ =\frac{-9}{-3}\\ =3\end{array}$

To find the slope of line LJ , substitute $-3$ for $y_2$ , 0 for $y_1$ , $-3$ for $x_2$ and $-12$ for $x_1$ in the formula.

  $\begin{array}{c}{m}_3=\frac{\left(-3\right)-0}{\left(-3\right)-\left(-12\right)}\\ =\frac{-3}{9}\\ =\frac{-1}{3}\end{array}$

Now, calculate the product of slopes of lines $KL\text{ and }LJ$ .

  $\begin{array}{l}{m}_2\times m_3=3\times \left(\frac{-1}{3}\right)\\ =-1\end{array}$

Since, the product of slopes of line $KL\text{ and }LJ$ is $-1$ . So, these lines are perpendicular to each other. The angle between the lines is a right-angle triangle.

Hence, it is proved that $\Delta JKL$ is a right angled triangle.