Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 2 - Mixed Review Exercises: 7

Answer

$y=-3$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Point-Slope Form with the given point $(2,-3)$ and the slope perpendicular to $x=2.$ $\bf{\text{Solution Details:}}$ The equation $x=2$ defines a vertical line. Perpendicular to a vertical line is a horizontal line. All horizontal lines have a slope of $0.$ Hence, the needed line has the following characteristics: \begin{array}{l}\require{cancel} \text{Slope: }m=0 \\\text{Through: }(2,-3) .\end{array} Using $y-y_1=m(x-x_1)$ or the Point-Slope Form of linear equations, the equation of the line with the given conditions, \begin{array}{l}\require{cancel} y_1=-3 ,\\x_1=2 ,\\m=0 ,\end{array} is \begin{array}{l}\require{cancel} y-(-3)=0(x-2) \\\\ y+3=0 \\\\ y=-3 .\end{array}
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