Question

What is the standard form of the line that passes through the points (-4, 6) and (0, -7)?

Ax+4y= -7
B
4x+y=-7
13x + 4y = -28
D-4x+y=-28
30960995

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

`(\stackrel{x_1}{-4}~,~\stackrel{y_1}{6})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{-7}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{-7}-\stackrel{y1}{6}}}{\underset{\textit{\large run}} {\underset{x_2}{0}-\underset{x_1}{(-4)}}} \implies \cfrac{-7 -6}{0 +4} \implies \cfrac{ -13 }{ 4 } \implies - \cfrac{13}{4}`

`\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}{6}=\stackrel{m}{- \cfrac{13}{4}}(x-\stackrel{x_1}{(-4)}) \implies y -6 = - \cfrac{13}{4} ( x +4) \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{4}}{4(y-6)=4\left( - \cfrac{13}{4} ( x +4) \right)}\implies 4y-24=-13(x+4) \\\\\\ 4y-24=-13x-52\implies 13x+4y-24=-52\implies \boxed{13x+4y=-28}`