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, what is the value of b?

Answer 1

its -21 cus im asian and smart

Explanation:

Answer 2

`J(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-5})\qquad K(\stackrel{x_2}{-6}~,~\stackrel{y_2}{3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{3}-\stackrel{y1}{(-5)}}}{\underset{run} {\underset{x_2}{-6}-\underset{x_1}{(-4)}}}\implies \cfrac{3+5}{-6+4}\implies \cfrac{8}{-2}\implies -4`

`\begin{array}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-\stackrel{y_1}{(-5)}=\stackrel{m}{-4}(x-\stackrel{x_1}{(-4)})\implies y+5=-4(x+4)`

`y+5=-4x-16\implies y=-4x\underset{\stackrel{\uparrow }{b}}{-21} \impliedby \begin{array}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}`