Question

3. Write the equation of a line in slope-intercept form that passes through the following points: (3,-8) and (-12,2).

Answer 1

just an addition to the decent posting above

`(\stackrel{x_1}{3}~,~\stackrel{y_1}{-8})\qquad (\stackrel{x_2}{-12}~,~\stackrel{y_2}{2}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{2}-\stackrel{y1}{(-8)}}}{\underset{\textit{\large run}} {\underset{x_2}{-12}-\underset{x_1}{3}}} \implies \cfrac{2 +8}{-15} \implies \cfrac{ 10 }{ -15 } \implies - \cfrac{ 2 }{ 3 }`

`\begin{array} \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}{(-8)}=\stackrel{m}{- \cfrac{ 2 }{ 3 }}(x-\stackrel{x_1}{3}) \implies y +8 = - \cfrac{ 2 }{ 3 } ( x -3) \\\\\\ y+8=- \cfrac{ 2 }{ 3 }x+2\implies {\Large \begin{array}{llll} y=- \cfrac{ 2 }{ 3 }x-6 \end{array}}`

Answer 2

Answer: Y=-2/3x-6

Step-by-step explanation:

we subtract both y1 x1 and y2 x2:

2-(-8)

-12-3 =

10/-15 = -2/3 (Equivalent fraction)

y=-2/3x + b

we need to plug in the first point which is (3.-8)

-8=-2/3(3) the 3 cancels out and we are left with -8=-2

we subtract and get -6 which is our Y intercept