Math Question Algebra

Can someone tell me what I am doing wrong?

Find the equation of the line determined by each pair of points.
(6, -2) (3,2)

First step find the slope: 4/3
Second change into point-Slope Form: y -(-2) = 4/3 (x - 6)
Third step, distributed the 4/3
4/3x + - 2/2 = 4/3x + 1
Fourth step, subtract (-2) from each side.
y = 4/3x + 3

Please help me. I don't know what I am doing wrong. I'm sure I misunderstood something.

Find the equation of the line determined by each pair of points.
(6, -2) (3,2)

The slope is (2-(-2)) / (3-6) = 4/(-3) = -4/3

First, slope = -4/3 Note it is negative
Second, y -(-2) = -4/3 (x - 6)
Third, distribute -4/3 Only to the x and the -6 on the right side
y + 2 = -4/3x + 8
Fourth, subtract 2 (since it was addition)
y + 2 - (2) = -4/3x + 8 - (2)
y = -4/3x + 6

Now change to standard form.
Multiply through by 3 (the denominator)
(3)y = (3)-4/3x + (3)6
3y = -4x + 18
Add 4x to both sides
(4x) + 3y = (4x) + -4x + 18
4x + 3y = 18

slope = (-2 - 2)/(6 - 3) = -4/3
y - (-2) = -4/3(x - 6)
y + 2 = -4x/3 + 24/3
y = -4x/3 + 8 - 2
y = -4x/3 + 6
If you want to put into standard form,
y = -4x/3 + 6
3(y = -4x/3 + 6)
3y = -4x + 18
4x + 3y = 18