Given:
Points on the line (5,2) and (-3,-2)
The slope-interept form of an equation is written as
[tex]\begin{gathered} y=mx+b \\ \text{where} \\ m\text{ is the slope of the line} \\ b\text{ is the y-intercept} \end{gathered}[/tex]
Solve first for the slope
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ \text{where} \\ (x_1,y_1)\text{ and }(x_2,y_2)\text{ are the points in the line} \\ \\ (x_1,y_1)=(5,2) \\ (x_2,y_2)=(-3,-2) \\ \\ \text{Substitute and the slope is} \\ m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{-2-2}{-3-5} \\ m=\frac{-4}{-8} \\ m=\frac{1}{2} \\ \\ \text{The slope of the line is }\frac{1}{2} \end{gathered}[/tex]
Now that we have solved for the slope, use any of the two points to solve for the y-intercept. For this case, we will use (5,2) but using (-3,-2) will work just as well.
[tex]\begin{gathered} y=mx+b \\ \text{Substitute} \\ m=\frac{1}{2},x=5,\text{ and }y=2 \\ \\ y=mx+b \\ 2=(\frac{1}{2})(5)+b \\ 2=\frac{5}{2}+b \\ 2-\frac{5}{2}=b \\ \frac{4}{2}-\frac{5}{2}=b \\ b=-\frac{1}{2} \end{gathered}[/tex]
Putting it together, with m = 1/2 and b = -1/2, the equation of the line is
[tex]y=\frac{1}{2}x-\frac{1}{2}[/tex]
