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: Start with point - slope form

We are given the point - slope form of the line: $y - 1=\frac{1}{4}(x - 12)$

Step2: Eliminate the fraction

Multiply both sides of the equation by 4 to get rid of the fraction.
$4\times(y - 1)=4\times\frac{1}{4}(x - 12)$
Using the distributive property $a(b - c)=ab - ac$, we have:
$4y-4=x - 12$

Step3: Rearrange to standard form ($Ax + By = C$)

We want to get the $x$ and $y$ terms on one side and the constant on the other. First, move the $x$ term to the left and the constant term to the right.
Subtract $x$ from both sides: $4y-4 - x=- 12$
Add 4 to both sides: $4y - x=-12 + 4$
Simplify the right - hand side: $4y - x=-8$
Multiply both sides by - 1 to make the $x$ coefficient positive: $x - 4y = 8$

Answer:

A. $x - 4y = 8$