College Algebra (10th Edition)

Published by Pearson

Chapter 2 - Section 2.3 - Lines - 2.3 Assess Your Understanding - Page 179: 68

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). The line we are looking for is perpendicular to $y=2x-3$ (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: $y=-\frac{1}{2}x+b \\-2 = -\frac{1}{2}(1)=b \\-2 = -\frac{1}{2} + b \\-2+\frac{1}{2}=b \\-\frac{4}{2}+\frac{1}{2}=b \\-\frac{3}{2}=b$ Thus, the equation of the line is: $\color{blue}{y=-\frac{1}{2}x-\frac{3}{2}}$

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