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: 23


The equation of the line that passes through the point $\left( -6,4 \right)$ and is perpendicular to the line that has an x-intercept of $2$ and a y-intercept of 4 in slope–intercept form is $y=-\frac{1}{2}x+1$.

Work Step by Step

First find the equation of the line with x-intercept of $2$ and a y-intercept of 4. Then, this line will pass through $\left( 2,0 \right)\text{ and }\left( 0,-4 \right)$. 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 ${{x}_{1}}=2\text{ and }{{x}_{2}}=0$ in the above formula to get: $\begin{align} & m=\frac{f\left( 2 \right)-f\left( 0 \right)}{2-0} \\ & =\frac{0-\left( -4 \right)}{2-0} \\ & =\frac{4}{2} \\ & =2 \end{align}$ The slope of the perpendicular line is the negative reciprocal of $2$. Thus, the slope of the perpendicular line is $-\frac{1}{2}$. Use the point-slope formula $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ Substitute the value of the slope of the line $m=-\frac{1}{2}$ and point $\left( {{x}_{1}},{{y}_{1}} \right)=\left( -6,4 \right)$ in equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$. The obtained expression is: $\begin{align} & y-4=-\frac{1}{2}\cdot \left( x-\left( -6 \right) \right) \\ & y-4=-\frac{1}{2}\cdot \left( x+6 \right) \end{align}$ Use the distributive property to get: $\begin{align} & y-4=-\frac{1}{2}x-3 \\ & y=-\frac{1}{2}x+1 \end{align}$ Thus, the required equation of the line is $y=-\frac{1}{2}x+1$.
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