Question

the point - slope form of the equation of a line that passes through points (8, 4) and (0, 2) is $y - 4=\frac{1}{4}(x - 8)$. what is the slope - intercept form of the equation for this line?
\\(\circ\\) $y = \frac{1}{4}x - 12$
\\(\circ\\) $y = \frac{1}{4}x - 4$
\\(\circ\\) $y = \frac{1}{4}x + 2$
\\(\circ\\) $y = \frac{1}{4}x + 6$

Explanation:

Step1: Expand the point - slope form

We start with the point - slope form \(y - 4=\frac{1}{4}(x - 8)\). Using the distributive property \(a(b - c)=ab - ac\), where \(a = \frac{1}{4}\), \(b=x\) and \(c = 8\), we get \(y-4=\frac{1}{4}x-\frac{1}{4}\times8\).
Simplify \(\frac{1}{4}\times8 = 2\), so the equation becomes \(y - 4=\frac{1}{4}x-2\).

Step2: Solve for y

To get the slope - intercept form \(y=mx + b\) (where \(m\) is the slope and \(b\) is the y - intercept), we add 4 to both sides of the equation \(y - 4=\frac{1}{4}x-2\).
\(y=\frac{1}{4}x-2 + 4\).
Simplify \(-2 + 4=2\), so \(y=\frac{1}{4}x + 2\).

Answer:

\(y=\frac{1}{4}x + 2\) (corresponding to the option \(y=\frac{1}{4}x + 2\))