Question

the point - slope form of the equation of the line that passes through (-9, -2) and (1, 3) is $y - 3=\frac{1}{2}(x - 1)$. what is the slope - intercept form of the equation for this line?
$y=\frac{1}{2}x-\frac{7}{2}$ $y=\frac{1}{2}x + 2$
$y=\frac{1}{2}x-4$ $y=\frac{1}{2}x+\frac{5}{2}$

Explanation:

Step1: Expand the point - slope form

We start with the point - slope form \(y - 3=\frac{1}{2}(x - 1)\). Using the distributive property \(a(b - c)=ab - ac\), where \(a = \frac{1}{2}\), \(b=x\) and \(c = 1\), we get \(y-3=\frac{1}{2}x-\frac{1}{2}\).

Step2: Solve for y

To get the slope - intercept form \(y=mx + b\), we add 3 to both sides of the equation. First, we rewrite 3 as \(\frac{6}{2}\) to have a common denominator. So \(y=\frac{1}{2}x-\frac{1}{2}+3=\frac{1}{2}x-\frac{1}{2}+\frac{6}{2}\). Then we combine the constant terms: \(-\frac{1}{2}+\frac{6}{2}=\frac{5}{2}\). So the equation becomes \(y=\frac{1}{2}x+\frac{5}{2}\).

Answer:

\(y=\frac{1}{2}x+\frac{5}{2}\) (the option with this equation)