Question

What is the Standard Form of the line with x-intercept of 6 and y-intercept of 5?

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

$\bf \stackrel{\textit{x-intercept}}{(\stackrel{x_1}{6}~,~\stackrel{y_1}{0})}\qquad \stackrel{\textit{y-intercept}}{(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})} ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{5}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{6}}}\implies -\cfrac{5}{6}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{-\cfrac{5}{6}}(x-\stackrel{x_1}{6}) \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{6}}{6(y-0)=6\left(-\cfrac{5}{6}(x-6) \right)}\implies 6y=-5(x-6) \\\\\\ 6y=-5x+30 \implies 5x+6y=30$