## Algebra 2 Common Core

$y+6=-16(x+1)$
With the given points, $(-1,-6)$ and $(-2,10),$ then \begin{align*} y_1&= -6 ,\\y_2&= 10 ,\\x_1&= -1 ,\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 connecting the two given points is \begin{align*} m&= \dfrac{-6-10}{-1-(-2)} \\\\&= \dfrac{-6-10}{-1+2} \\\\&= \dfrac{-16}{1} \\\\&= -16 .\end{align*} Using $y-y_1=m(x-x_1)$ or the Point-Slope Form, with $m= -16$ and using the point $(-1,-6)$, then \begin{align*} y-(-6)&=-16(x-(-1)) \\ y+6&=-16(x+1) .\end{align*} A Point-Slope Form of the line with the given conditions is $y+6=-16(x+1) .$