Chapter 4.6, Problem 17PSC

To determine

To show: The median from $C\text{ to }\overline{AB}$ is not longer than the altitude from $C\text{ to }\overline{AB}$ .

Explanation of Solution

Given information:

  $\Delta ABC\text{ has vertices at }A=\left(2,1\right),B=\left(12,3\right),\text{ and }C=\left(6,7\right)$ .

Calculation:

Given $\Delta ABC\text{ has vertices at }A=\left(2,1\right),B=\left(12,3\right),\text{ and }C=\left(6,7\right)$ .

Drawing the triangle as below.

  

In above triangle midpoint D from $C\text{ to }\overline{AB}$ is $\left(7,2\right)$ .

  $\begin{array}{l}\text{Slope of }\overline{AB}=\frac{3-1}{12-2}=\frac{1}{5}\\ \text{Slope of }\overline{CD}=\frac{2-7}{7-6}=-5\\ \text{Slope of }\overline{CP}=-\frac{1}{\text{slope of}\overline{AB}}=-\frac{1}{\frac{1}{5}}=-5\\ \text{Since Slope of }\overline{CD}=\text{Slope of }\overline{CP}\\ \text{and both line have a common point C}\text{. This means both lines are same}\text{. }\\ \text{Therefore median from }C\text{ to }\overline{AB}\text{ and altitude from }C\text{ to }\overline{AB}\text{ are equal}\text{.}\\ \text{In other words, the median from }C\text{ to }\overline{AB}\text{ is not longer than altitude from }C\text{ to }\overline{AB}.\\ \end{array}$