Question

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

Answer 1

$m=\dfrac{y_2-y_1}{x_2-x_1}\\\\m=\dfrac{1-(-3)}{12-(-4)}=\dfrac{4}{16}=\dfrac{1}{4}$

Correct equation of the line is

$y-1=\dfrac{1}{4}(x-12)$

The standard form is: Ax + By = C.

$y-1=\dfrac{1}{4}(x-12)\qquad|\text{use distributive property}\\\\y-1=\dfrac{1}{4}x-3\qquad|\text{multiply both sides by 4}\\\\4y-4=x-12\qquad|\text{add 12 to both sides}\\\\4y+8=x\qquad|\text{subtract 4y from both sides}\\\\8=x-4y$

Answer: x - 4y = 8