Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 8 - Section 8.1 - Review of Equations of Lines and Writing Parallel and Perpendicular Lines - Practice: 1

Answer

Practice 1 (Solution) Since the slope and a point on the line are given, the point-slope form can be used, with m = –2 and (x1, y1) = (2, –4). y – y1 = m (x – x1) Point-slope form y – (–4) = –2 (x – 2) Substitute m = –2 and (x1, y1) = (2, –4) y + 4 = –2x + 4 Using the distributive property y = –2x In slope-intercept form, the equation is y = –2x.

Work Step by Step

Practice 1 (Solution) Since the slope and a point on the line are given, the point-slope form can be used, with m = –2 and (x1, y1) = (2, –4). y – y1 = m (x – x1) Point-slope form y – (–4) = –2 (x – 2) Substitute m = –2 and (x1, y1) = (2, –4) y + 4 = –2x + 4 Using the distributive property y = –2x In slope-intercept form, the equation is y = –2x.
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