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

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 3 - Introduction to Graphing - 3.7 Point-Slope Form and Equations of Lines - 3.7 Exercise Set - Page 218: 28



Work Step by Step

RECALL: (1) The point-slope form of a line's equation is $y-y_1=m(x-x_1)$ where $m$=slope and $(x_1, y_1)$ is a point on the line. (2) The slope-intercept form of a line's equation is $y=mx+b$ where $m$=slope and $b$ is the y-coordinate of the line's y-intercept. (3) Parallel lines have equal slopes. (4) Perpendicular lines have slopes whose product is $-1$ (negative reciprocals of each other). Write $4x+7y=1$ in slope-intercept form: $4x+7y=1 \\7y=1-4x \\7y=-4x+1 \\\frac{7y}{7}=\frac{-4x+1}{7} \\y=-\frac{4}{7}x+\frac{1}{7}$ This means the equation $4x+7y=1$ is equivalent to $y=-\frac{4}{7}x+\frac{1}{7}$. The line is perpendicular to $y=-\frac{4}{7}x+\frac{1}{7}$. Since the slope of this line is $-\frac{4}{7}$, then the slope of the line perpendicular to it is the negative reciprocal of $-\frac{4}{7}$, which is $\frac{7}{4}=1.75$. Using the given point on the line $(0, -4.2)$ (which is the y-intercept) and the slope $1.75$, the slope-intercept form of the line's equation is: $y=1.75x+(-4.2) \\y=1.75x-4.2$
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