coordinate geometry

Line t: $y= \frac{-3}{4}x+\frac{5}{4}$

Perpendicular also means the "negative reciprocal of the slope" and in this case the negative reciprocal of the slope of Line t is $\frac{4}{3}$. 

Line p: $y=\frac{4}{3}x+?$

To find the y-intercept, we need to put in the point (5, 5) into Line p.

$5=\frac{4}{3}(5)+?$ ---> $5=\frac{20}{3}+?$---> $5-\frac{20}{3}=?$---> $\frac{-5}{3}=?$

Line p: $y=\frac{4}{3}x+\frac{-5}{3}$

To find the point common to both lines, we set them equal to each other:

$\frac{4}{3}x+\frac{-5}{3}$=$\frac{-3}{4}x+\frac{5}{4}$
$x=\frac{7}{5}$

Since we only need the "x" coordinate, $x=\frac{7}{5}$ is our answer.

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