## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 3 - Introduction to Graphing - 3.7 Point-Slope Form and Equations of Lines - 3.7 Exercise Set - Page 218: 58

#### Answer

$y=\dfrac{4}{5}(x-6)$

#### Work Step by Step

Using the properties of equality, the given linear equation, $5x+4y=1$ is equivalent to \begin{array}{l} 4y=-5x+1 \\\\ y=-\dfrac{5}{4}x+\dfrac{1}{4} .\end{array} Using $y=mx+b$ or the Slope-Intercept form where $m$ is the slope, then the slope of the given line is $-\dfrac{5}{4}$. Since perpendicular lines have negative reciprocal slopes, then the needed linear equation has slope equal to $\dfrac{4}{5}$. Since it also passes through the given point $( 6,0 )$, then using $y-y_1=m(x-x_1)$ or the Point-Slope form where $m$ is the slope and $(x_1,y_1)$ is a point on the line, the equation of the needed line is \begin{array}{l} y-0=\dfrac{4}{5}(x-6) \\\\ y=\dfrac{4}{5}(x-6) .\end{array}

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