Question

Write an equation in standard form of the line passing through points

Answer 1

$\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -\frac{4}{9} &,& \frac{4}{9}~) % (c,d) &&(~ \frac{6}{5} &,& \frac{4}{5}~) \end{array} \\\\\\ % slope = m slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{\frac{4}{5}-\frac{4}{9}}{\frac{6}{5}-\left( -\frac{4}{9} \right)}\implies \cfrac{\frac{4}{5}-\frac{4}{9}}{\frac{6}{5}+\frac{4}{9}}$

$\bf \cfrac{\quad \frac{36-20}{45}\quad }{\frac{54+20}{45}}\implies \cfrac{\quad\frac{16}{45} \quad }{\frac{74}{45}}\implies \cfrac{16}{45}\cdot \cfrac{45}{74}\implies \cfrac{16}{74}\implies \cfrac{8}{37} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-\cfrac{4}{9}=\cfrac{8}{37}\left[ x -\left( -\cfrac{4}{9} \right)\right]$

$\bf y-\cfrac{4}{9}=\cfrac{8}{37}\left( x +\cfrac{4}{9} \right)\implies y-\cfrac{4}{9}=\cfrac{8}{37}x+\cfrac{32}{333} \\\\\\ y=\cfrac{8}{37}x+\cfrac{32}{333}+\cfrac{4}{9}\implies y=\cfrac{8}{37}x+\cfrac{32}{333}+\cfrac{148}{333}\implies y=\cfrac{8}{37}x+\cfrac{180}{333} \\\\\\ y=\cfrac{8}{37}x+\cfrac{20}{37}\implies \stackrel{standard~form}{-\cfrac{8}{37}x+y=\cfrac{20}{37}}$