Question

Write an equation in standard form of the line passing through the points (1, 3) and (-5, 6)

Answer 1

keeping in mind that 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

$\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-5}~,~\stackrel{y_2}{6}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{6}-\stackrel{y1}{3}}}{\underset{run} {\underset{x_2}{-5}-\underset{x_1}{1}}}\implies \cfrac{3}{-6}\implies -\cfrac{1}{2}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{-\cfrac{1}{2}}(x-\stackrel{x_1}{1})\implies y-3=-\cfrac{1}{2}x+\cfrac{1}{2} \\\\\\ y=-\cfrac{1}{2}x+\cfrac{1}{2}+3\implies y=-\cfrac{1}{2}x+\cfrac{7}{2}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{2}}{2(y)=2\left( -\cfrac{1}{2}x+\cfrac{7}{2} \right)} \\\\\\ 2y=-x+7\implies x+2y=7$