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

The hhpotenuse of a right angled triangle has its ends at the points (1,3) and (−4,1) . Find an equation of the legs (perpendicar sides) of the triangle.

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

Answered by MJS last updated on 04/Dec/19

given hypothenuse ⇒ the 3^(rd) point is located on a circle with center in the center of the hyp. and radius=((hyp.)/2) center=(( ((1),(3) )+ (((−4)),(1) ))/2)= (((−(3/2))),(2) ) radius=(1/2)abs ( ((1),(3) )− (((−4)),(1) ))=(1/2)(√(29)) equation of circle is (x+(3/2))^2 +(y−2)^2 =((29)/4) y=2±(√(−x^2 −3x+5)) ⇒ −(3/2)−((√(29))/2)≤x≤−(3/2)+((√(29))/2) so there′s no unique triangle triangles ABC_1 , ABC_2 let A= (((−4)),(1) ) B= ((1),(3) ) C_(1, 2) = ((p),((2±(√(−p^2 −3p+5)))) ) equations of lines AB: y=(2/5)x+((13)/5) AC_1 : y=((1−(√(−p^2 −3p+5)))/(p+4))x+((p+8−4(√(−p^2 −3p+5)))/(p+4)) BC_1 : y=−((1+(√(−p^2 −3p+5)))/(p−1))x+((3p−2+(√(−p^2 −3p+5)))/(p−1)) AC_2 : y=((1+(√(−p^2 −3p+5)))/(p+4))x+((p+8+4(√(−p^2 −3p+5)))/(p+4)) BC_2 : y=−((1−(√(−p^2 −3p+5)))/(p−1))x+((3p−2−(√(−p^2 −3p+5)))/(p−1)) ∀p∈R∣−(3/2)−((√(29))/2)≤x≤−(3/2)+((√(29))/2)∧p≠1∧p≠−4

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

Commented by vishalbhardwaj last updated on 04/Dec/19

sir equation of perpendicular sides ??

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

Commented by MJS last updated on 04/Dec/19

ran out of time, will finish in about an hour or two

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

Commented by MJS last updated on 04/Dec/19

is this what you need?

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

Commented by vishalbhardwaj last updated on 04/Dec/19

yes sir

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

Commented by vishalbhardwaj last updated on 05/Dec/19

but sir the equations of perpendicular lines are given x = 1 and y = 1, why ??

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

Commented by MJS last updated on 05/Dec/19

I don′t know. the triangle is not unique, we would need more information anyway the point ((1),(1) ) lies on the circle, p=1 ⇒ we cannot use my formulas but we can calculate the lines AC and BC and yes, they are x=1 and y=1 with the given information this is only one solution

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

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