Chapter 13, Problem 11CR

To determine

To find slope of the line $\overline{\text{QT}}$ and using it find the slope of line altitude to $\overline{\text{QT}}$

Answer to Problem 11CR

  $\frac{3}{4},\frac{-4}{3}$

Explanation of Solution

Given Points are

  $\text{Q(-1,-2) , T(7,4)}$

Slope of a line passing through $(x_1,y_1),(x_2,y_2)$ is

Slope , $\text{m = }\frac{y_2-y_1}{x_2-x_1}$

Here,

Let $(x_1,y_1)=\text{(-1,-2) , }(x_2,y_2)=\text{(7,4)}$

Therefore, Slope of $\overline{\text{QT}}\text{, m = }\frac{y_2-y_1}{x_2-x_1}=\frac{4-(-2)}{7-(-1)}$

  $\text{m = }\frac{4+2}{7+1}=\frac{6}{8}=\frac{3}{4}$

Now, Slope of altitude of $\overline{\text{QT}}\text{ = }\frac{-1}{\text{slope of }\overline{\text{QT}}}$

Slope of altitude of $\overline{\text{QT}}\text{ = }\frac{-1}{\frac{3}{4}}=\frac{-4}{3}$