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Question Number 74948 by vishalbhardwaj last updated on 04/Dec/19

$$\mathrm{The}\mathrm{hhpotenuse}\mathrm{of}\mathrm{a}\mathrm{right}\mathrm{angled}\mathrm{triangle}$$$$\mathrm{has}\mathrm{its}\mathrm{ends}\mathrm{at}\mathrm{the}\mathrm{points}(1,3)\mathrm{and}(-4,1)$$$$.\mathrm{Find}\mathrm{an}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{legs}(\mathrm{perpendicar}$$$$\mathrm{sides})\mathrm{of}\mathrm{the}\mathrm{triangle}.$$

Answered by MJS last updated on 04/Dec/19

$$\mathrm{given}\mathrm{hypothenuse}\Rightarrow \mathrm{the}{3}^{\mathrm{rd}}\mathrm{point}\mathrm{is}\mathrm{located}$$$$\mathrm{on}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{center}\mathrm{in}\mathrm{the}\mathrm{center}\mathrm{of}\mathrm{the}$$$$\mathrm{hyp}.\mathrm{and}\mathrm{radius}=\frac{\mathrm{hyp}.}{2}$$$$\mathrm{center}=\frac{\left(\begin{array}{c}1\\ 3\end{array}\right)+\left(\begin{array}{c}-4\\ 1\end{array}\right)}{2}=\left(\begin{array}{c}-\frac{3}{2}\\ 2\end{array}\right)$$$$\mathrm{radius}=\frac{1}{2}\mathrm{abs}(\left(\begin{array}{c}1\\ 3\end{array}\right)-\left(\begin{array}{c}-4\\ 1\end{array}\right))=\frac{1}{2}\sqrt{29}$$$$\mathrm{equation}\mathrm{of}\mathrm{circle}\mathrm{is}$$$${(x+\frac{3}{2})}^2+{(y-2)}^2=\frac{29}{4}$$$$y=2\pm \sqrt{-x^2-3x+5}\Rightarrow -\frac{3}{2}-\frac{\sqrt{29}}{2}⩽x⩽-\frac{3}{2}+\frac{\sqrt{29}}{2}$$$$\mathrm{so}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{unique}\mathrm{triangle}$$$$$$$$\mathrm{triangles}{ABC}_1,{ABC}_2$$$$\mathrm{let}A=\left(\begin{array}{c}-4\\ 1\end{array}\right)B=\left(\begin{array}{c}1\\ 3\end{array}\right)C_{1,2}=\left(\begin{array}{c}p\\ 2\pm \sqrt{-p^2-3p+5}\end{array}\right)$$$$\mathrm{equations}\mathrm{of}\mathrm{lines}$$$$AB:y=\frac{2}{5}x+\frac{13}{5}$$$${AC}_1:y=\frac{1-\sqrt{-p^2-3p+5}}{p+4}x+\frac{p+8-4\sqrt{-p^2-3p+5}}{p+4}$$$${BC}_1:y=-\frac{1+\sqrt{-p^2-3p+5}}{p-1}x+\frac{3p-2+\sqrt{-p^2-3p+5}}{p-1}$$$${AC}_2:y=\frac{1+\sqrt{-p^2-3p+5}}{p+4}x+\frac{p+8+4\sqrt{-p^2-3p+5}}{p+4}$$$${BC}_2:y=-\frac{1-\sqrt{-p^2-3p+5}}{p-1}x+\frac{3p-2-\sqrt{-p^2-3p+5}}{p-1}$$$$\mathrm{\forall }p\in \mathbb{R}\mid -\frac{3}{2}-\frac{\sqrt{29}}{2}⩽x⩽-\frac{3}{2}+\frac{\sqrt{29}}{2}\wedge p\ne 1\wedge p\ne -4$$$$$$

Commented by vishalbhardwaj last updated on 04/Dec/19

$$\mathrm{sir}\mathrm{equation}\mathrm{of}\mathrm{perpendicular}\mathrm{sides}??$$

Commented by MJS last updated on 04/Dec/19

$$\mathrm{ran}\mathrm{out}\mathrm{of}\mathrm{time},\mathrm{will}\mathrm{finish}\mathrm{in}\mathrm{about}\mathrm{an}\mathrm{hour}$$$$\mathrm{or}\mathrm{two}$$

Commented by MJS last updated on 04/Dec/19

$$\mathrm{is}\mathrm{this}\mathrm{what}\mathrm{you}\mathrm{need}?$$

Commented by vishalbhardwaj last updated on 04/Dec/19

$$\mathrm{yes}\mathrm{sir}$$

Commented by vishalbhardwaj last updated on 05/Dec/19

$$\mathrm{but}\mathrm{sir}\mathrm{the}\mathrm{equations}\mathrm{of}\mathrm{perpendicular}$$$$\mathrm{lines}\mathrm{are}\mathrm{given}x=1\mathrm{and}y=1,\mathrm{why}??$$

Commented by MJS last updated on 05/Dec/19

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}.\mathrm{the}\mathrm{triangle}\mathrm{is}\mathrm{not}\mathrm{unique},\mathrm{we}$$$$\mathrm{would}\mathrm{need}\mathrm{more}\mathrm{information}$$$$\mathrm{anyway}\mathrm{the}\mathrm{point}\left(\begin{array}{c}1\\ 1\end{array}\right)\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{circle},p=1$$$$\Rightarrow \mathrm{we}\mathrm{cannot}\mathrm{use}\mathrm{my}\mathrm{formulas}\mathrm{but}\mathrm{we}\mathrm{can}$$$$\mathrm{calculate}\mathrm{the}\mathrm{lines}AC\mathrm{and}BC\mathrm{and}\mathrm{yes},\mathrm{they}$$$$\mathrm{are}x=1\mathrm{and}y=1$$$$\mathrm{with}\mathrm{the}\mathrm{given}\mathrm{information}\mathrm{this}\mathrm{is}\mathrm{only}\mathrm{one}$$$$\mathrm{solution}$$

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