Chapter 7.7, Problem 31E

To determine

To find: the lengths of altitudes in triangle with vertices. $A,B\text{ and }C$ .

Answer to Problem 31E

  $\frac{34}{5},\frac{17\sqrt{26}}{26}\text{ and }\frac{34\sqrt{53}}{53}$

Explanation of Solution

Given information:

  $A\left(-3,4\right),B\left(1,7\right)\text{ and }C\left(-1,-3\right)$

Calculation:

Evaluate the equation for line $AB$ as shown below.

  $\begin{array}{c}y-y_B=\frac{y_B-y_A}{x_B-x_A}\left(x-x_B\right)\\ y-7=\frac{7-4}{1-\left(-3\right)}\left(x-1\right)\left[\text{Put the point coordinates}\right]\\ y-7=\frac{3}{4}x-\frac{3}{4}\\ 4y-28=3x-3\left[\text{Multiply by 4}\right]\\ -3x+4y-25=0\end{array}$

The distance to point $C\left(-1,-3\right)$ can be evaluated as follows: $\begin{array}{c}d=\left|\frac{A{x}_1+B{y}_1+C}{\sqrt{A^2+B^2}}\right|\\ =\left|\frac{\left(-3\right)\left(-15\right)+\left(4\right)\left(-3\right)+\left(-25\right)}{\sqrt{{\left(-3\right)}^2+{\left(4\right)}^2}}\right|\left[\begin{array}{l}\text{Put }A=-3,B=4,C=-25,\\ x_1\text{=}-1\text{ and }{y}_1=-3\end{array}\right]\\ =\frac{-34}{5}\end{array}$

Thus, the first altitude is $\frac{-34}{5}$ .

Evaluate the equation for line $BC$ as shown below.

  $\begin{array}{c}y-y_B=\frac{y_C-y_B}{x_C-x_B}\left(x-x_B\right)\\ y-7=\frac{-3-7}{-1-1}\left(x-1\right)\left[\text{Put the point coordinates}\right]\\ y-7=5x-5\\ 5x-y+2=0\end{array}$

The distance to point $A\left(-3,4\right)$ can be evaluated as follows: $\begin{array}{c}d=\left|\frac{A{x}_1+B{y}_1+C}{\pm \sqrt{A^2+B^2}}\right|\\ =\left|\frac{\left(5\right)\left(-3\right)+\left(-1\right)\left(4\right)+\left(2\right)}{-\sqrt{{\left(5\right)}^2+{\left(-1\right)}^2}}\right|\left[\begin{array}{l}\text{Put }A=5,B=-1,C=2,\\ x_1\text{=}-\text{3 and }{y}_1=4\end{array}\right]\\ =\frac{17}{\sqrt{26}}\\ =\frac{17\sqrt{26}}{26}\left[\text{Multiply and divide by }\sqrt{26}\right]\end{array}$

Thus, the second altitude is $\frac{17\sqrt{26}}{26}$ .

Evaluate the equation for line $AC$ as shown below.

  $\begin{array}{c}y-y_C=\frac{y_C-y_A}{x_C-x_A}\left(x-x_C\right)\\ y-\left(-3\right)=\frac{-3-4}{-1-\left(-3\right)}\left(x-\left(-1\right)\right)\left[\text{Put the point coordinates}\right]\\ y+3=-\frac{7}{2}x-\frac{7}{2}\\ 2y+6=-7x-7\left[\text{Multiply by 2}\right]\\ 7x+2y+13=0\end{array}$

The distance to point $B\left(1,7\right)$ can be evaluated as follows: $\begin{array}{c}d=\left|\frac{A{x}_1+B{y}_1+C}{\pm \sqrt{A^2+B^2}}\right|\\ =\left|\frac{\left(7\right)\left(1\right)+\left(2\right)\left(7\right)+\left(13\right)}{-\sqrt{{\left(7\right)}^2+{\left(2\right)}^2}}\right|\left[\begin{array}{l}\text{Put }A=7,B=2,C=13,\\ x_1\text{=}1\text{ and }{y}_1=7\end{array}\right]\\ =\frac{34}{-\sqrt{53}}\\ =\frac{34\sqrt{53}}{53}\left[\text{Multiply and divide by }-\sqrt{53}\right]\end{array}$

Thus, the third altitude is $\frac{34\sqrt{53}}{53}$ .