The point-slope form of the equation of the line that passes through (–4, –3) and (12, 1) is y – 1 = (x – 12). what is the standard form of the equation for this line?

x – 4y = 8
x – 4y = 2
4x – y = 8
4x – y = 2

(-4,-3) , (12,1)
slope = (1 - (-3) / (12 - (-4) = 4/16 = 1/4

y - y1 = m(x - x1)
slope(m) = 1/4
(12,1)...x = 12 and y = 1
sub
y - 1 = 1/4(x - 12)
y - 1 = 1/4x - 3
y = 1/4x - 3 + 1
y = 1/4x - 2
-1/4x + y = -2
x - 4y = 8 <=== standard form