## Algebra 2 Common Core

$y+3=\dfrac{5}{3}(x+4)$
With the given points, $(-4,-3)\text{ and } (2,7),$ then \begin{align*} y_1&= -3 ,\\y_2&= 7 ,\\x_1&= -4 ,\text{ and }\\ x_2&= 2 .\end{align*} Using $m=\dfrac{y_1-y_2}{x_1-x_2}$ or the Slope Formula, the slope, $m,$ of the line is \begin{align*}\require{cancel} m&=\dfrac{-3-7}{-4-2} \\\\&= \dfrac{-10}{-6} \\\\&= \dfrac{\cancel{-10}^5}{\cancel{-6}^3} \\\\&= \dfrac{5}{3} .\end{align*} Using $y-y_1=m(x-x_1)$ or the Point-Slope Form of linear equations, with the point $(-4,-3)$ and $m=\dfrac{5}{3},$ then \begin{align*} y-(-3)&=\dfrac{5}{3}(x-(-4)) \\\\ y+3&=\dfrac{5}{3}(x+4) .\end{align*} Hence, an equation of the line in Point-Slope Form is $y+3=\dfrac{5}{3}(x+4) .$