Respuesta :
parallel lines, have the same slope, so the slope of the line through 0,0 and -2,-12, is the same as for the line running through (6/5,-19.5) as well, so what is it anyway?
[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 0}}\quad ,&{{ 0}})\quad % (c,d) &({{ -2}}\quad ,&{{ -12}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{-12-0}{-2-0}\implies \cfrac{-12}{-2}\implies 6[/tex]
so, we're looking for the equation of a line whose slope is 6, and goes through (6/5,-19/5)
[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ \frac{6}{5}}}\quad ,&{{ -\frac{19}{5}}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies 6 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-\left( -\frac{19}{5} \right)=6\left(x-\frac{6}{5} \right) \\\\\\ y+\cfrac{19}{5}=6x-\cfrac{36}{5}\implies y=6x-\cfrac{36}{5}-\cfrac{19}{5}\implies y=6x-\cfrac{55}{5} \\\\\\ y=6x-11[/tex]
[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 0}}\quad ,&{{ 0}})\quad % (c,d) &({{ -2}}\quad ,&{{ -12}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{-12-0}{-2-0}\implies \cfrac{-12}{-2}\implies 6[/tex]
so, we're looking for the equation of a line whose slope is 6, and goes through (6/5,-19/5)
[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ \frac{6}{5}}}\quad ,&{{ -\frac{19}{5}}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies 6 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-\left( -\frac{19}{5} \right)=6\left(x-\frac{6}{5} \right) \\\\\\ y+\cfrac{19}{5}=6x-\cfrac{36}{5}\implies y=6x-\cfrac{36}{5}-\cfrac{19}{5}\implies y=6x-\cfrac{55}{5} \\\\\\ y=6x-11[/tex]
