Question

a line passes through (3,-2) and (6,2). write an equation in point-slope form. rewrite the equation in standard form ​

Answer 1

again, bearing 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

$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{-2})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{2}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{2-(-2)}{6-3}\implies \cfrac{2+2}{6-3}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-2)=\cfrac{4}{3}(x-3)\implies y+2=\cfrac{4}{3}x-4 \\\\[-0.35em] ~\dotfill$

$\bf y=\cfrac{4}{3}x-6\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3(y)=3\left( \cfrac{4}{3}x-6 \right)}\implies 3y=4x-18 \\\\\\ -4x+3y=-18\implies \stackrel{\textit{standard form}}{4x-3y=18}$