Chapter 7.6, Problem 32E

To determine

To find: the lengthof the incircle.

Answer to Problem 32E

6 units

Explanation of Solution

Given information:

Incircle is a unique circle on which three sides of a triangle are tangent.

The center of incircle of $\Delta ABC$ is at origin.

The line equations are as follows:

  $\begin{array}{c}x+4y=6\sqrt{17}\\ 2x+\sqrt{5}y=-18\\ 2\sqrt{2}x=y+18\end{array}$

Calculation:

Evaluate any line equation in normal form, i.e. $x\cos \varphi +y\sin \varphi -p=0$ . The length of incircle $p$ can be evaluated as follows:

  $\begin{array}{l}2\sqrt{2}x=y+18\left[\text{Subtract }\left(y+18\right)\text{ from each side}\right]\\ 2\sqrt{2}x-y-18=0\left[\text{Here }A=2\sqrt{2}\text{, }B=-1\text{ and }C=18\right]\\ \frac{2\sqrt{2}}{\sqrt{{\left(2\sqrt{2}\right)}^2+{\left(-1\right)}^2}}x-\frac{y}{\sqrt{{\left(2\sqrt{2}\right)}^2+{\left(-1\right)}^2}}-\frac{18}{\sqrt{{\left(2\sqrt{2}\right)}^2+{\left(-1\right)}^2}}=0\left[\begin{array}{l}\text{}\text{Divide each side by}\\ -\sqrt{A^2+B^2}\end{array}\right]\\ \frac{2\sqrt{2}}{\sqrt{9}}x-\frac{y}{\sqrt{9}}-\frac{18}{\sqrt{9}}=0\\ \frac{2\sqrt{2}}{3}x-\frac{y}{3}-6=0\left[\text{Normal line equation}\right]\end{array}$

Thus, the length of the incircle $\left(p\right)$ is 6 units.