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?

      y-y1 = m (x-x1)          y+3 = (1/4)(x+4)                 [using point (x1,y1)]      or          y–1 = (1/4)(x–12)                [using point (x2,y2]   [let's use this one since you did] To put this into Standard Form (Ax+By=C),          y–1 = (1/4)(x–12)           4y - 4 = x - 12          -x + 4y - 4 = -12           [subtract x from both sides]          -x + 4y = -8                 [add 4 to both sides]           x - 4y = 8            [multiply by (-1) to have x-coefficient positive