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e-lub [12.9K]
1 year ago
7

The points (3,-10) and (7,r) lie on a line with slope 5/4. Find the missing coordinate r.

Mathematics
1 answer:
LenaWriter [7]1 year ago
5 0

(\stackrel{x_1}{3}~,~\stackrel{y_1}{-10})\qquad (\stackrel{x_2}{7}~,~\stackrel{y_2}{r}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{r}-\stackrel{y1}{(-10)}}}{\underset{run} {\underset{x_2}{7}-\underset{x_1}{3}}} \implies \cfrac{r +10}{4}~~ = ~~\stackrel{\stackrel{m}{\downarrow }}{\cfrac{5}{4}} \\\\\\ 4r+40=20\implies 4r=-20\implies r=\cfrac{-20}{4}\implies {\Large \begin{array}{llll} r=-5 \end{array}}

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\left\{\begin{array}{ccc}5x-y=5&|subtract\ 5x\ from\ both\ sides\\5x-3y=15\end{array}\right\\\left\{\begin{array}{ccc}-y=-5x+5&|change\ the\ signs-multiply\ both\ sides\ by\ (-1)\\5x-3y=15\end{array}\right\\\left\{\begin{array}{ccc}y=5x-5&(1^o)\\5x-3y=15&(2^o)\end{array}\right\\\\subtitute\ (1^o)\ to\ (2^o)\\\\5x-3(5x-5)=15\\5x-3(5x)-3(-5)=15\\5x-15x+15=15\\-10x+15=15\ \ \ \ |subtract\ 15\ from\ both\ sides\\-10x=0\ \ \ \ \ |divide\ both\ sides\ by\ (-10)\\\boxed{x=0}

subtitute\ value\ of\ x\ to\ (1^o):\\\\y=5(0)-5=0-5=\boxed{-5}\\\\Answer:\boxed{\boxed{\left\{\begin{array}{ccc}x=0\\y=-5\end{array}\right}}
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