Question

I am trying to find the equation of the following lines, i’m just having trouble putting them in slope-intercept form.

slope= 2/5 through the point (-1,-6)

through the points (-2,5) and (5,8)

Answer 1

`\bf \begin{array}{ccccccccc} &&x_1&&y_1\\ &&(~ -1 &,& -6~) \end{array} \\\\\\ slope = m\implies \cfrac{2}{5} \\\\\\ \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-6)=\cfrac{2}{5}[x-(-1)] \\\\\\ y+6=\cfrac{2}{5}(x+1)\implies y+6=\cfrac{2}{5}x+\cfrac{2}{5} \\\\\\ y=\cfrac{2}{5}x+\cfrac{2}{5}-6\implies y=\cfrac{2}{5}x-\cfrac{28}{5}`




`\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ &&(~ -2 &,& 5~) &&(~ 5 &,& 8~) \end{array} \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{8-5}{5-(-2)}\implies \cfrac{8-5}{5+2}\implies \cfrac{3}{7} \\\\\\ \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-5=\cfrac{3}{7}[x-(-2)] \\\\\\ y-5=\cfrac{3}{7}(x+2)\implies y-5=\cfrac{3}{7}x+\cfrac{6}{7} \\\\\\ y=\cfrac{3}{7}x+\cfrac{6}{7}+5\implies y=\cfrac{3}{7}x+\cfrac{41}{7}`