Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 3 - Parallel and Perpendicular Lines - 3-8 Slopes of Parallel and Perpendicular Lines - Practice and Problem-Solving Exercises - Page 204: 53


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

Work Step by Step

We are given the points $A(0, 3)$ and $B(6, 0)$. Let's use the formula to find the slope $m$ given two points: $m = \frac{y_2 - y_1}{x_2 - x_1}$, where $(x_1, y_1)$ and $(x_2, y_2)$ are two points on the line. Let's plug in these values into this formula: $m = \frac{0 - 3}{6 - 0}$ Subtract the numerator and denominator to simplify: $m = \frac{-3}{6}$ Divide the numerator and denominator by their greatest common denominator, which is $3$: $m = -\frac{1}{2}$ Now that we have the slope, we can use one of the points and plug these values into the point-slope equation, which is given by the formula: $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is a point on the line. Let's plug in the points and slope into the formula: $y - 0 = -\frac{1}{2}(x - 6)$ Let's use the distributive property: $y = -\frac{1}{2}x - (\frac{1}{2})(-6)$ Multiply to simplify: $y = -\frac{1}{2}x + \frac{6}{2}$ Divide both the numerator and denominator of $\frac{6}{2}$ by their greatest common factor, $2$: $y = -\frac{1}{2}x + 3$
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