## Algebra: A Combined Approach (4th Edition)

Practice 2 (Solution) The slope of the line, m, is m = $\frac{(4-0)}{(-2-3)}$ = $\frac{-4}{5}$ Making use of the point-slope form and replace (x1, y1) by (3, 0) and m = $\frac{-4}{5}$, y – y1 = m (x – x1) Point-slope form y – 0 = $\frac{-4}{5}$(x – 3) Substitute (x1, y1) by (3, 0) and m = $\frac{-4}{5}$ 5y = –4 (x – 3) Simplify 5y = –4x + 12 4x + 5y = 12 In standard form, the equation is 4x + 5y = 12.
Practice 2 (Solution) The slope of the line, m, is m = $\frac{(4-0)}{(-2-3)}$ = $\frac{-4}{5}$ Making use of the point-slope form and replace (x1, y1) by (3, 0) and m = $\frac{-4}{5}$, y – y1 = m (x – x1) Point-slope form y – 0 = $\frac{-4}{5}$(x – 3) Substitute (x1, y1) by (3, 0) and m = $\frac{-4}{5}$ 5y = –4 (x – 3) Simplify 5y = –4x + 12 4x + 5y = 12 In standard form, the equation is 4x + 5y = 12.