Let line t be the line represented by 3x + 4y = 5 and let line p be the line perpendicular to line t and containing the point (5, 5). What is the x-coordinate of the point common to line t and line p ?
+934 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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