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Are-give-two-parallel-lines-and-a-point-A-is-given-between-them-and-the-distance-from-A-to-the-lines-are-a-and-b-Determine-the-cathetus-of-right-angled-triangle-in-A-knowing-that-the-other-vertices

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Question Number 76386 by Maclaurin Stickker last updated on 27/Dec/19
Are give two parallel lines and a point A is given between them, and the distance from A to the lines are a and b. Determine the cathetus of right−angled triangle in A knowing that the other vertices belong to the parallel lines and the area of the triangle is equal to k^2 .
$${Are}\:{give}\:{two}\:{parallel}\:{lines}\:{and}\:{a}\:{point} \\ $$$${A}\:{is}\:{given}\:{between}\:{them},\:{and}\:{the} \\ $$$${distance}\:{from}\:{A}\:{to}\:{the}\:{lines}\:{are}\: \\ $$$$\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}.\:{Determine}\:{the}\:{cathetus}\:{of} \\ $$$${right}−{angled}\:{triangle}\:{in}\:{A}\:{knowing} \\ $$$${that}\:{the}\:{other}\:{vertices}\:{belong}\:{to} \\ $$$${the}\:{parallel}\:{lines}\:{and}\:{the}\:{area}\:{of} \\ $$$${the}\:{triangle}\:{is}\:{equal}\:{to}\:\boldsymbol{{k}}^{\mathrm{2}} . \\ $$
Commented by Maclaurin Stickker last updated on 27/Dec/19
thank you
$${thank}\:{you} \\ $$
Commented by MJS last updated on 27/Dec/19
no. the distance between a point and a line is the minimum distance ⇒ a and b are part of a line right angled to the 2 parallels
$$\mathrm{no}.\:\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{a}\:\mathrm{point}\:\mathrm{and}\:\mathrm{a}\:\mathrm{line} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{distance}\:\Rightarrow\:{a}\:\mathrm{and}\:{b}\:\mathrm{are}\:\mathrm{part} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{line}\:\mathrm{right}\:\mathrm{angled}\:\mathrm{to}\:\mathrm{the}\:\mathrm{2}\:\mathrm{parallels} \\ $$
Answered by mr W last updated on 27/Dec/19
(x/a)=(b/y) ⇒y=((ab)/x) AC=(√(a^2 +x^2 ))=a(√(1+(x^2 /a^2 ))) AB=(√(b^2 +y^2 ))=(√(b^2 +((a^2 b^2 )/x^2 )))=b(√(1+(a^2 /x^2 ))) (1/2)(√((a^2 +x^2 )(b^2 +((a^2 b^2 )/x^2 ))))=k^2 =area (1+(x^2 /a^2 ))(1+(a^2 /x^2 ))=(((2k^2 )/(ab)))^2 let ξ=(x^2 /a^2 ), λ=(((2k^2 )/(ab)))^2 (1+ξ)(1+(1/ξ))=λ ξ^2 −(λ−2)ξ+1=0 ⇒ξ=((λ−2±(√(λ(λ−4))))/2) ⇒x=a(√((λ−2±(√(λ(λ−4))))/2)) we see there is always solution(s) if λ≥4, i.e. k^2 ≥ab. catheti of the searched triangle: AC=a(√(1+ξ)) AB=b(√(1+(1/ξ)))
$$\frac{{x}}{{a}}=\frac{{b}}{{y}} \\ $$$$\Rightarrow{y}=\frac{{ab}}{{x}} \\ $$$${AC}=\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }={a}\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }} \\ $$$${AB}=\sqrt{{b}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\sqrt{{b}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}={b}\sqrt{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{x}^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)\left({b}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)}={k}^{\mathrm{2}} ={area} \\ $$$$\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)=\left(\frac{\mathrm{2}{k}^{\mathrm{2}} }{{ab}}\right)^{\mathrm{2}} \\ $$$${let}\:\xi=\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} },\:\lambda=\left(\frac{\mathrm{2}{k}^{\mathrm{2}} }{{ab}}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{1}+\xi\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\xi}\right)=\lambda \\ $$$$\xi^{\mathrm{2}} −\left(\lambda−\mathrm{2}\right)\xi+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\xi=\frac{\lambda−\mathrm{2}\pm\sqrt{\lambda\left(\lambda−\mathrm{4}\right)}}{\mathrm{2}} \\ $$$$\Rightarrow{x}={a}\sqrt{\frac{\lambda−\mathrm{2}\pm\sqrt{\lambda\left(\lambda−\mathrm{4}\right)}}{\mathrm{2}}} \\ $$$${we}\:{see}\:{there}\:{is}\:{always}\:{solution}\left({s}\right) \\ $$$${if}\:\lambda\geqslant\mathrm{4},\:{i}.{e}.\:{k}^{\mathrm{2}} \geqslant{ab}. \\ $$$$ \\ $$$${catheti}\:{of}\:{the}\:{searched}\:{triangle}:\: \\ $$$${AC}={a}\sqrt{\mathrm{1}+\xi} \\ $$$${AB}={b}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\xi}} \\ $$
Commented by mr W last updated on 27/Dec/19
Commented by john santu last updated on 27/Dec/19
great sir
$${great}\:{sir} \\ $$
Commented by john santu last updated on 27/Dec/19
sir why ∡A = 90^o ?
$${sir}\:{why}\:\measuredangle{A}\:=\:\mathrm{90}^{{o}} ? \\ $$
Commented by mr W last updated on 27/Dec/19
it is given in the question that the right angle is at point A.
$${it}\:{is}\:{given}\:{in}\:{the}\:{question}\:{that}\:{the} \\ $$$${right}\:{angle}\:{is}\:{at}\:{point}\:{A}. \\ $$
Commented by john santu last updated on 27/Dec/19
ohh yes. thanks you sir
$${ohh}\:{yes}.\:{thanks}\:{you}\:{sir} \\ $$
Commented by john santu last updated on 27/Dec/19
how tu drawing this triangle in your phone sir?
$${how}\:{tu}\:{drawing}\:{this}\:{triangle}\: \\ $$$${in}\:{your}\:{phone}\:{sir}? \\ $$
Commented by mr W last updated on 27/Dec/19
i don′t use any special app. i just use the app which is preinstalled in my huawei smart phone.
$${i}\:{don}'{t}\:{use}\:{any}\:{special}\:{app}.\:\:{i}\:{just} \\ $$$${use}\:{the}\:{app}\:{which}\:{is}\:{preinstalled} \\ $$$${in}\:{my}\:{huawei}\:{smart}\:{phone}. \\ $$

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