Respuesta :

[tex]\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ 3 &,& -4~) % (c,d) &&(~ 5 &,& 1~) \end{array} \\\\\\ % slope = m slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-(-4)}{5-3}\implies \cfrac{1+4}{5-3}\implies \cfrac{5}{2} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-4)=\cfrac{5}{2}(x-3)\implies y+4=\cfrac{5}{2}x-\cfrac{15}{2}[/tex]

[tex]\bf y=\cfrac{5}{2}x-\cfrac{15}{2}-4\implies y=\cfrac{5}{2}x-\cfrac{23}{2}\impliedby \begin{array}{llll} \textit{now let's multiply both}\\ \textit{sides by }\stackrel{LCD}{2} \end{array} \\\\\\ 2(y)=2\left( \cfrac{5}{2}x-\cfrac{23}{2} \right)\implies 2y=5x-23\implies \stackrel{standard~form}{-5x+2y=-23} \\\\\\ \textit{and if we multiply both sides by -1}\qquad 5x-2y=23[/tex]

side note:  multiplying by the LCD of both sides is just to get rid of the denominators
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