Chapter 13.3, Problem 7WE

To determine

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

Answer to Problem 7WE

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

Explanation of Solution

Given:

Coordinates of vertices of $\Delta ABC$ are $A\left(0,0\right),B\left(7,3\right),C\left(2,-5\right)$ .

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 $A\left(0,0\right),B\left(7,3\right),C\left(2,-5\right)$

Slope of line of $\overline{AB}$

  $m_{AB}=\frac{3-0}{7-0}=\frac{3}{7}$

Slope of line of $\overline{AC}$

  $m_{AC}=\frac{-5-0}{2-0}=\frac{-5}{2}$

Slope of line of $\overline{BC}$

  $m_{BC}=\frac{-5-3}{2-7}=\frac{-8}{-5}=\frac{8}{5}$

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 $AB$

  $\begin{array}{l}{m}_{CF}\times m_{AB}=-1\\ m_{CF}=-\frac{1}{m_{AB}}=-\frac{1}{\left(\frac{3}{7}\right)}=-\frac{7}{3}\end{array}$

altitude $BD$ is perpendicular to side $AC$

  $\begin{array}{l}{m}_{BD}\times m_{AC}=-1\\ m_{BD}=-\frac{1}{m_{AC}}=-\frac{1}{\left(-\frac{5}{2}\right)}=\frac{2}{5}\end{array}$

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