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: \(y - 1=\frac{1}{4}(x - 12)\)
First, we want to eliminate the fraction. Multiply both sides of the equation by 4 to get rid of the denominator.
\(4(y - 1)=4\times\frac{1}{4}(x - 12)\)
Simplify the right - hand side: \(4(y - 1)=x - 12\)

Step2: Distribute the 4 on the left - hand side

Using the distributive property \(a(b - c)=ab - ac\), where \(a = 4\), \(b=y\) and \(c = 1\), we have:
\(4y-4=x - 12\)

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

We want to move all the \(x\) and \(y\) terms to one side and the constant to the other side. Let's add 4 to both sides and subtract \(x\) from both sides.
\(-x + 4y=-12 + 4\)
Simplify the right - hand side: \(-x + 4y=-8\)
Multiply both sides of the equation by - 1 to make the coefficient of \(x\) positive:
\(x-4y = 8\) Wait, no, wait. Wait, let's go back. Wait, when we had \(4y-4=x - 12\), let's move \(x\) to the left and the constant to the right.
Subtract \(x\) from both sides: \(-x + 4y-4=-12\)
Add 4 to both sides: \(-x + 4y=-8\)
Multiply both sides by - 1: \(x - 4y = 8\)

Answer:

\(x - 4y = 8\) (corresponding to the option "x - 4y = 8")