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Write an equation of the line containing the point and parallel to the given line.

(6,-4);4x-3y=7

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


4x-3y=7\implies 4x=7+3y\implies 4x-7=3y \\\\\\ \cfrac{4x-7}{3}=y\implies \stackrel{\stackrel{m}{\downarrow }}{\cfrac{4}{3}}x-\cfrac{7}{3}=y\qquad \impliedby \qquad \begin{array} \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 are really looking for the equation of a line whose slope is 4/3 and it passes through (6 , -4)


(\stackrel{x_1}{6}~,~\stackrel{y_1}{-4})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{4}{3} \\\\\\ \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}{(-4)}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{6}) \implies y +4 = \cfrac{4}{3} ( x -6) \\\\\\ y+4=\cfrac{4}{3} x-8\implies {\Large \begin{array}{llll} y=\cfrac{4}{3}x-12 \end{array}}

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