Question

the point - slope form of the equation of the line that passes through (-4, -3) and (12, 1) is ( y - 1=\frac{1}{4}(x - 12) ). what is the standard form of the equation for this line?
( \bigcirc x - 4y = 8 )
( \bigcirc x - 4y = 2 )
( \bigcirc 4x - y = 8 )
( \bigcirc 4x - y = 2 )

Explanation:

Step1: Eliminate the fraction

Multiply both sides by 4:
$$4(y - 1) = 4 \times \frac{1}{4}(x - 12)$$
$$4y - 4 = x - 12$$

Step2: Rearrange to standard form

Move all terms to one side to get $Ax + By = C$:
$$-x + 4y = -12 + 4$$
$$-x + 4y = -8$$
Multiply by -1 to make the x-coefficient positive:
$$x - 4y = 8$$

Answer:

x - 4y = 8 (Option A)