Chapter 8 - Section 8.1 - Review of Equations of Lines and Writing Parallel and Perpendicular Lines - Practice - Page 567: 1

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.

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.