Respuesta :
[tex]\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ -7 &,& -5~)
% (c,d)
&&(~ -5 &,& 4~)
\end{array}
\\\\\\
% slope = m
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{4-(-5)}{-5-(-7)}\implies \cfrac{4+5}{-5+7}\implies \cfrac{9}{2}[/tex]
[tex]\bf \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-5)=\cfrac{9}{2}[x-(-7)] \\\\\\ \boxed{y+5=\cfrac{9}{2}(x+7)}\implies y+5=\cfrac{9}{2}x+\cfrac{63}{2}\impliedby \begin{array}{llll} \textit{and now we multiply}\\ \textit{both sides by }\stackrel{LCD}{2} \end{array} \\\\\\ 2y+10=9x+63\implies \boxed{\stackrel{\textit{standard form}}{-9x+2y=53}}[/tex]
just a quick note, multiplying both sides by the LCD, whatever it may be, simply gets rid of the denominators.
[tex]\bf \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-5)=\cfrac{9}{2}[x-(-7)] \\\\\\ \boxed{y+5=\cfrac{9}{2}(x+7)}\implies y+5=\cfrac{9}{2}x+\cfrac{63}{2}\impliedby \begin{array}{llll} \textit{and now we multiply}\\ \textit{both sides by }\stackrel{LCD}{2} \end{array} \\\\\\ 2y+10=9x+63\implies \boxed{\stackrel{\textit{standard form}}{-9x+2y=53}}[/tex]
just a quick note, multiplying both sides by the LCD, whatever it may be, simply gets rid of the denominators.
