Question

What is the equation of the line that passes through the points (6,8) and (-3,2)

Answer 1

The point-slope form:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (6, 8) and (-3, 2). Substitute:

$m=\dfrac{2-8}{-3-6}=\dfrac{-6}{-9}=\dfrac{2}{3}\\\\y-8=\dfrac{2}{3}(x-6)\qquad|\text{use distributive property}\\\\y-8=\dfrac{2}{3}x-4\qquad|+8\\\\y=\dfrac{2}{3}x+4\qquad|-\dfrac{2}{3}x\\\\-\dfrac{2}{3}x+y=4\qquad|\cdot(-3)\\\\2x-3y=-12$

Answer:

point-slope form: $y-8=\dfrac{2}{3}(x-6)$

slope-intercept form: $y=\dfrac{2}{3}x+4$

standard form: $2x-3y=-12$