Question

Find the equation of the line passing through point (-4,-5) and is parallel to the line 7x-5y=-6

Answer 1

keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above

`7x-5y=-6\implies -5y=-7x-6\implies y=\cfrac{-7x-6}{-5} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{7}{5}}x+\cfrac{6}{5}\qquad \impliedby \qquad \begin{array}ll \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}`

so we're really looking for the equation of a line whose slope is 7/5 and it passes through (-4 , -5)

`(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-5})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{7}{5} \\\\\\ \begin{array}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-\stackrel{y_1}{(-5)}=\stackrel{m}{ \cfrac{7}{5}}(x-\stackrel{x_1}{(-4)}) \implies y +5 = \cfrac{7}{5} ( x +4) \\\\\\ y+5=\cfrac{7}{5}x+\cfrac{28}{5}\implies y=\cfrac{7}{5}x+\cfrac{28}{5}-5\implies {\Large \begin{array}{llll} y=\cfrac{7}{5}x+\cfrac{3}{5} \end{array}}`