## 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: 30

#### Answer

$y=x+27$

#### 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 $x-y=6$ in slope-intercept form: $x-y=6 \\-y=6-x \\-1(-y)=-1(6-x) \\y=-6+x \\y=x-6$ This means the equation $x-y=6$ is equivalent to $y=x-6$. The line is parallel to $y=x-6$. Since the slope of this line is $1$, then the slope of the line parallel to it is also $1$. Using the given point on the line $(0, 27)$ (which is the y-intercept) and the slope $1$, the equation of the line in point-slope form is: $y=x+27$

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