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?\
$x - 4y = 2$\
$4x - y = 8$\
$4x - y = 2$\
$x - 4y = 8$

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. According to the distributive property \(a(b + c)=ab+ac\), we have:
\(4(y - 1)=4\times\frac{1}{4}(x - 12)\)
Simplify both sides: \(4y-4=x - 12\)

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

Subtract \(4y\) from both sides and add 12 to both sides.
\(4y-4-4y + 12=x - 12-4y + 12\)
Simplify the equation: \(x-4y=8\)

Answer:

\(x - 4y = 8\)