## College Algebra 7th Edition

$y=-8x-11$
RECALL: (1) The slope-intercept form of a line's equation is $y=mx+b$ where m = slope and b = y-intercept. (2) The slope of a line can be found using the formula $m=\dfrac{y_2-y_1}{x_2-x_1}$ where $(x_1, y_1)$ and $(x_2, y_2)$ are points on the line. Solve for the slope to obtain: $m=\dfrac{-3-5}{-1-(-2)} \\m=\dfrac{-8}{-1+2} \\m=\dfrac{-8}{1} \\m=-8$ Thus, the tentative equation of the line is: $y=-8x+b$ To find the value of $b$, substitute the coordinates of $(-2, 5)$ into the tentative equation above to obtain: $y=-8x+b \\5 = -8(-2)+ b \\5 = 16+b \\5-16=b \\-11=b$ Therefore, the equation of the line is: $y=-8x+(-11) \\y=-8x-11$