Re: Give the right triangle ABC, find th slope of the segment
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09 Apr 2024, 08:57
OE
B. This problem tests your understanding of the proportions on a right triangle. Because of the orientation of the triangle, you can determine the rise for each segment. The missing part of the slope is the run, which corresponds to the height of the triangle (the unlabeled segment that intersects the hypotenuse at a right angle). Fortunately, this value can be found using the ratios of the proportions of a right triangle. Let the height of the triangle be h.
Use the ratios of the hypotenuses to find h:
short leg of the small triangle/hypotenuse of the small triangle=short leg of the large triangle/hypotenuse of the large triangle
\(\frac{2}{AB}=\frac{AB}{5}\)
\(10=AB^2\)
\(2^2+h^2=AB^2\)
\(h=\sqrt{6}\)
Now
slope \(\overline{AB}=\dfrac{2}{\sqrt{6}}=\dfrac{\sqrt{6}}{3}\)
slope \(\overline{BC}=\dfrac{-3}{\sqrt{6}}=\dfrac{- \sqrt{6}}{2}\)