Answer
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.
Work Step by Step
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.
