Question

Equation in standard form using the given points

Answer 1

standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

now, to get the equation of any straight line, we simply need two points off of it, let's use those two in the picture below.

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

`\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}{6}=\stackrel{m}{- \cfrac{4 }{ 3 }}(x-\stackrel{x_1}{(-6)}) \implies y -6 = - \cfrac{4 }{ 3 } ( x +6) \\\\\\ \stackrel{\textit{multiplying both sides by}\stackrel{LCD}{3}}{3(y-6)=3\left( - \cfrac{4 }{ 3 } ( x +6) \right)} \implies 3y-18=-4(x+6) \\\\\\ 3y-18=-4x-24\implies 3y=-4x-6\implies {\Large \begin{array}{llll} 4x+3y=-6 \end{array}}`