## 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: 25

#### Answer

$y=\dfrac{4}{3}x-12$

#### 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). The line is perpendicular to $y=−\frac{3}{4}x+1$. Since the slope of this line is $−\frac{3}{4}$, then the slope of the line perpendicular to it is the negative reciprocal of $−\frac{3}{4}$, which is $\frac{4}{3}$. Using the given point on the line $(0, −12)$ and the slope $\frac{4}{3}$, the equation of the line in slope-intercept form is: $y=\dfrac{4}{3}x+(-12) \\y=\dfrac{4}{3}x-12$

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