## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 2 - Section 2.5 - Zeros of Polynomial Functions - Exercise Set - Page 380: 99

#### Answer

The equation of the perpendicular line is $\underline{y-3=5\left( x+5 \right);5x-y+28=0}$.

#### Work Step by Step

Consider the equation of the straight line as follows. \begin{align} & x+5y-7=0 \\ & y=\left( -\frac{1}{5} \right)x+\frac{7}{5} \end{align} The slope of this line is ${{m}_{1}}=-\frac{1}{5}$ . Let the slope of the line perpendicular to this line be ${{m}_{2}}$. Then,   \begin{align} & {{m}_{1}}{{m}_{2}}=-1 \\ & \left( -\frac{1}{5} \right){{m}_{2}}=-1 \\ & {{m}_{2}}=5 \end{align} Let $\left( x,y \right)$ be any point on the perpendicular line passing through $\left( -5,3 \right)$. \begin{align} & \frac{y-3}{x+5}=5 \\ & y-3=5\left( x+5 \right) \end{align} Therefore, the equation of the perpendicular line in point-slope form is: $y-3=5\left( x+5 \right)$ And, in general form it is as follows: $5x-y+28=0$ Therefore, the equation of the perpendicular line is $y-3=5\left( x+5 \right);5x-y+28=0$.

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