choose the equation that represents a line that passes through points -3,2 and 2,15x+y=-135x-y=17x-5y=-13x+5y=7​

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[tex]\bf (\stackrel{x_1}{-3}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-2}{2-(-3)}\implies \cfrac{1-2}{2+3}\implies -\cfrac{1}{5} \\\\\\ \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{1}{5}[x-(-3)]\implies y-2=-\cfrac{1}{5}(x+3)[/tex][tex]\bf y-2=-\cfrac{1}{5}x-\cfrac{3}{5}\implies y=-\cfrac{1}{5}x-\cfrac{3}{5}+2\implies y=-\cfrac{1}{5}x+\cfrac{7}{5} \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{5}}{5(y)=5\left( -\cfrac{1}{5}x+\cfrac{7}{5} \right)}\implies 5y=-x+7\implies \blacktriangleright x+5y=7 \blacktriangleleft[/tex]