Using the two - point from
y − y1y2 − y1= x − x1x2 − x1\;\frac{y\;-\;y_1}{y_2\;-\;y_1} = \;\frac{x\;-\;x_1}{x_2\;-\;x_1}y2−y1y−y1=x2−x1x−x1
y − 2−2 − 2= x − 4−8 − 4\;\frac{y\;-\;2}{-2\;-\;2} = \;\frac{x\;-\;4}{-8\;-\;4}−2−2y−2=−8−4x−4
y − 2−4= x − 4−12\;\frac{y\;-\;2}{-4} = \;\frac{x\;-\;4}{-12}−4y−2=−12x−4
−12(y −2)−4\;\frac{-12(y\;-2)}{-4}−4−12(y−2) = x - 4
3(y -2) = x -4
3y - 6 = x - 4
3y = x - 4 + 6
3y = x + 2...
By comparing the equations;
3y = px + , p = 1
