## College Algebra (10th Edition)

Published by Pearson

# Chapter 2 - Section 2.3 - Lines - 2.3 Assess Your Understanding - Page 179: 69

#### Answer

$\color{blue}{y=\frac{1}{2}x+\frac{3}{2}}$

#### Work Step by Step

RECALL: (1) The slope-intercept form of a line's equation is: $y=mx+b$ where $m$ = slope and $b$ = y-intercept (2) Perpendicular lines have slopes whose product is $-1$ (negative reciprocals of each other). Write the equation of the given line in slope-intercept form to obtain: $2x+y=2 \\y=-2x+2$ The line we are looking for is perpendicular to the line above (whose slope is $-2$). This means that the line has a slope of $\frac{1}{2}$ (since $\frac{1}{2}(-2)=-1$). Thus, a tentative equation of the line we are looking for is: $y=\frac{1}{2}x+b$ To find the value of $b$, substitute the x and y values of the given point to obtain: $y=\frac{1}{2}x+b \\0 = \frac{1}{2}(-3)=b \\0 = -\frac{3}{2} + b \\0+\frac{3}{2}=b \\\frac{3}{2}=b$ Thus, the equation of the line is: $\color{blue}{y=\frac{1}{2}x+\frac{3}{2}}$

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