Algebra Equation please help

A line passes through the points (-2,-3) and (6,9). Find its equation in standard form.
_x+2y=_

Slope =(9--3)/(6--2)=12/8=3/2
so 3x-2y=c
Plug in(6,9)
18-18=0
then 3x-2y= 0
or
-3x+2y=0

Standard form is Ax + By = C, so we need to find the slope between the two points which is also equal to -A/B and then we can plug in a coordinate.

m = [9-(-3)]/[6-(-2)]
= (9+3)/(6+2)
= 12/8
= 3/2

if 3/2 = -A/B and B=2 then A=-3

so we have -3x + 2y = C

(-2, -3)

-3(-2) + 2(-3) = C
6-6=C
0=C

2y-3x=0 is your answer.

line passes through the points (-2,-3) and (6,9)
.'.equation: (y-y1)/(y1-y2) = (x-x1)/(x1-x2)
or, (y+3)/(-12) = (x+2)/(-8)
or, 8(y+3) = 12(x+2)
or, 2(y+3) = 3(x+2)
or, 2y - 3x = 0
Gap1: -3
Gap2: 0

Use point-slope form
m=3/2
y+3=(3/2)(x+2)
y=3x/2