Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.5 - More on Slope - Exercise Set - Page 226: 24


The equation of the line that passes through the point $\left( -5,6 \right)$ and is perpendicular to the line that has an x-intercept of $3$ and a y-intercept of $-9$ in slope–intercept form is $y=-\frac{1}{3}x+\frac{13}{3}$.

Work Step by Step

First find the equation of the line with an x-intercept of $3$ and a y-intercept of $-9$. This line will pass through $\left( 3,0 \right)\text{ and }\left( 0,-9 \right)$. Use these points to find the slope as: $m=\frac{f\left( {{x}_{2}} \right)-f\left( {{x}_{1}} \right)}{{{x}_{2}}-{{x}_{1}}}$ Put the value of ${{x}_{1}}=3\text{ and }{{x}_{2}}=0$ in the above formula to get: $\begin{align} & m=\frac{f\left( 3 \right)-f\left( 0 \right)}{3-0} \\ & =\frac{0-\left( -9 \right)}{3-0} \\ & =\frac{9}{3} \\ & =3 \end{align}$ The slope of the perpendicular line is the negative reciprocal of $3$. Thus, the slope of the perpendicular line is $-\frac{1}{3}$. Use the point-slope formula $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ Substitute the value of the slope of the line $m=-\frac{1}{3}$ and point $\left( {{x}_{1}},{{y}_{1}} \right)=\left( -5,6 \right)$ in equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$. The obtained expression is: $\begin{align} & y-6=-\frac{1}{3}\cdot \left( x-\left( -5 \right) \right) \\ & y-6=-\frac{1}{3}\cdot \left( x+5 \right) \end{align}$ Use the distributive property to get: $\begin{align} & y-6=-\frac{1}{3}x-\frac{5}{3} \\ & y=-\frac{1}{3}x-\frac{5}{3}+6 \\ & y=-\frac{1}{3}x+\frac{13}{3} \end{align}$ Thus, the required equation of the line is $y=-\frac{1}{3}x+\frac{13}{3}$.
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