Question

Line JK passes through points J(-4,-5) and K(-6,3). If the equation of the line is written in slope-intercept form, y = mx + b

Answer 1

Answer:

–21

Step-by-step explanation:

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Answer 2

$\bf J(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-5})\qquad K(\stackrel{x_2}{-6}~,~\stackrel{y_2}{3}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{3-(-5)}{-6-(-4)}\implies \cfrac{3+5}{-6+4}\implies \cfrac{8}{-2}\implies -4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-5)=-4[x-(-4)]\implies y+5=-4(x+4)$

$\bf y+5=-4x-16\implies y=-4x\stackrel{b}{\boxed{-21}}\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$