Chapter 13.3, Problem 8WE

To determine

To find the slope of each side and each altitude of $\Delta ABC$ .

Answer to Problem 8WE

The slope of line $\overline{AB}$ , $\overline{AC}$ and $\overline{BC}$ are $\frac{7}{2},-3,-\frac{8}{5}$ and the slope of altitudes $AE$ , $CF$ and $BD$ are $\frac{5}{8},-\frac{2}{7},\frac{1}{3}$ .

Explanation of Solution

Given:

Coordinates of vertices of $\Delta ABC$ are $A\left(1,4\right),B\left(-1,-3\right),C\left(4,-5\right)$ .

Calculation:

Below is the figure of $\Delta ABC$ and $AE$ , $BD$ and $CF$ are the altitudes.

  

Slope of a line when two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are given is

  $m=\frac{y_2-y_1}{x_2-x_1}$

Coordinates of vertices of $\Delta ABC$ are $A\left(1,4\right),B\left(-1,-3\right),C\left(4,-5\right)$

Slope of line of $\overline{AB}$

  $m_{AB}=\frac{-3-4}{-1-1}=\frac{-7}{-2}=\frac{7}{2}$

Slope of line of $\overline{AC}$

  $m_{AC}=\frac{-5-4}{4-1}=\frac{-9}{3}=-3$

Slope of line of $\overline{BC}$

  $m_{BC}=\frac{-5-3}{4-\left(-1\right)}=\frac{-8}{5}$

Since, altitude $AE$ is perpendicular to side $BC$

  $\begin{array}{l}{m}_{AE}\times m_{BC}=-1\\ m_{AE}=-\frac{1}{m_{BC}}=-\frac{1}{\left(-\frac{8}{5}\right)}=\frac{5}{8}\end{array}$

altitude $CF$ is perpendicular to side $AB$

  $\begin{array}{l}{m}_{CF}\times m_{AB}=-1\\ m_{CF}=-\frac{1}{m_{AB}}=-\frac{1}{\left(\frac{7}{2}\right)}=-\frac{2}{7}\end{array}$

altitude $BD$ is perpendicular to side $AC$

  $\begin{array}{l}{m}_{BD}\times m_{AC}=-1\\ m_{BD}=-\frac{1}{m_{AC}}=-\frac{1}{\left(-3\right)}=\frac{1}{3}\end{array}$

Therefore, the slope of line $\overline{AB}$ , $\overline{AC}$ and $\overline{BC}$ are $\frac{7}{2},-3,-\frac{8}{5}$ and the slope of altitudes $AE$ , $CF$ and $BD$ are $\frac{5}{8},-\frac{2}{7},\frac{1}{3}$ .