Chapter 4.8, Problem 32HP

To determine

Tocalculate:

The coordinates of point $L$ right triangle.

Answer to Problem 32HP

The coordinates of point $L$ is $\left(2a,0\right)$ or $\left(0,2b\right)$ .

Explanation of Solution

Given information:

Indicated triangle $=△JKL$

Coordinates:

  $\begin{array}{l}J\left(0,0\right)\\ K\left(2a,2b\right)\end{array}$

Formula Used:

Slope formula:

  $m=\frac{y_2-y}{x_2-x_1}$

Calculation:

First draw right triangle from point $K$ to the $x-$ axis to find a point $L$ that would make $△JKL$ a right triangle.

The vertical segment from $K$ would intersect the $x-$ axis at $\left(2a,0\right)$ , so use the slopes of the sides to verify that $△JKL$ is a right triangle.

  

Now use the slope formula:

  $m=\frac{y_2-y}{x_2-x_1}$

Put the coordinates as $\left(x_{1,}{y}_1\right)=\left(0,0\right),\left(x_{2,}{y}_2\right)=\left(2a,0\right)$ for $JL$

Simplify,

  $\begin{array}{l}m\overline{JL}=\frac{2a-0}{0-0}\\ m\overline{JL}=\frac{2a}{0}\end{array}$

Put the coordinates as $\left(x_{1,}{y}_1\right)=\left(2a,0\right),\left(x_{2,}{y}_2\right)=\left(2a,2b\right)$ for $KL$

Simplify,

  $\begin{array}{l}m\overline{KL}=\frac{2a-2a}{2b-0}\\ m\overline{KL}=\frac{0}{2b}or0\end{array}$

Since, the slope are undefined and $0,\overline{JL}\perp \overline{KL}$ and $△JKL$ a right triangle. The horizontal segment would intersect the $y-$ axis at $\left(0,2b\right)$ .

The slopes will again confirm that $\overline{JL}\perp \overline{KL}$ , so this point would also make $△JKL$ a right triangle.

Thus, the coordinates of $L$ could be either $\left(2a,0\right)$ or $\left(0,2b\right)$ .