Question

A line passes through the point (-6,3) and a slope of 5/2

Answer 1

Step-by-step explanation:

The point-slope form of an equation of a line:

$y-y_1=m(x-x_1)$

m - slope

(x₁, y₁) - point on a line

We have

$(-6,\ 3)\to x_1=-6,\ y_1=3,\ m=\dfrac{5}{2}$

Substitute:

$y-3=\dfrac{5}{2}(x-(-6))\\\\y-3=\dfrac{5}{2}(x+6)$

Convert to the slope-intercept form

$y=mx+b$

m - slope

b - y-intercept

$y-3=\dfrac{5}{2}(x+6)$         use the distributive property

$y-3=\dfrac{5}{2}x+\dfrac{5}{2\!\!\!\!\diagup_1}\cdot6\!\!\!\!\diagup^3$

$y-3=\dfrac{5}{2}x+15$           add 3 to both sides

$y=\dfrac{5}{2}x+18$

Convert to the standard form:

$Ax+By=C$

$y=\dfrac{5}{2}x+18$              multiply both sides by 2

$2y=5x+36$              subtract 5x from both sides

$-5x+2y=36$          change the signs

$5x-2y=-36$