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Question Number 43005 by mondodotto@gmail.com last updated on 06/Sep/18
The base of triangle passes through  a fixed point p(a,b) and its sides  are respectively bisected at right angles   by the line x+y=0 and x=9y  if the locus of the third vartex is a circle.  then find its equation.
$$\mathrm{The}\:\mathrm{base}\:\mathrm{of}\:\mathrm{triangle}\:\mathrm{passes}\:\mathrm{through} \\ $$$$\mathrm{a}\:\mathrm{fixed}\:\mathrm{point}\:\mathrm{p}\left(\mathrm{a},\mathrm{b}\right)\:\mathrm{and}\:\mathrm{its}\:\mathrm{sides} \\ $$$$\mathrm{are}\:\mathrm{respectively}\:\mathrm{bisected}\:\mathrm{at}\:\mathrm{right}\:\mathrm{angles}\: \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{line}\:\mathrm{x}+\mathrm{y}=\mathrm{0}\:\mathrm{and}\:\mathrm{x}=\mathrm{9y} \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{third}\:\mathrm{vartex}\:\mathrm{is}\:\mathrm{a}\:\mathrm{circle}. \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{its}\:\mathrm{equation}. \\ $$
Commented by mondodotto@gmail.com last updated on 06/Sep/18
please help me this problem
$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{this}\:\mathrm{problem} \\ $$
Commented by ajfour last updated on 06/Sep/18
Commented by ajfour last updated on 06/Sep/18
let A(h,k)  ⇒  B(−k, −h)  let C(α, β)  reflection of C in y=mx+c   is  α=h−((2m(c+mh−k))/(1+m^2 ))  β = k+((2(c+mh−k))/(1+m^2 ))  here we have  y= (x/9)  ⇒  m=(1/9) , c=0  ⇒  α = h−(((2/9)((h/9)−k))/(1+(1/(81))))  ⇒  𝛂=h−(9/(41))((h/9)−k)  and   𝛃 = k+((81)/(41))((h/9)−k)  eq. of line through BC is  y+h = ((β+h)/(α+k))(x+k)  As this passes through P (a,b)  b+h = ((β+h)/(α+k))(a+k)  replacing α, β in terms of h,k  and changing h,k to x,y we get  the desired locus of A(third vertex)  b+h=((k+((81)/(41))((h/9)−k)+h)/(h−(9/(41))((h/9)−k)+k))(a+k)  ⇒ (b+x)[x+y−(9/(41))((x/9)−y)]               = (a+y)[x+y+((81)/(41))((x/9)−y)]  locus  found is     40(x^2 +y^2 )−41(b−a)(x+y)                        −9(ax−by)=0 .
$${let}\:{A}\left({h},{k}\right) \\ $$$$\Rightarrow\:\:{B}\left(−{k},\:−{h}\right) \\ $$$${let}\:{C}\left(\alpha,\:\beta\right) \\ $$$${reflection}\:{of}\:{C}\:{in}\:{y}={mx}+{c}\:\:\:{is} \\ $$$$\alpha={h}−\frac{\mathrm{2}{m}\left({c}+{mh}−{k}\right)}{\mathrm{1}+{m}^{\mathrm{2}} } \\ $$$$\beta\:=\:{k}+\frac{\mathrm{2}\left({c}+{mh}−{k}\right)}{\mathrm{1}+{m}^{\mathrm{2}} } \\ $$$${here}\:{we}\:{have}\:\:\boldsymbol{{y}}=\:\frac{\boldsymbol{{x}}}{\mathrm{9}} \\ $$$$\Rightarrow\:\:{m}=\frac{\mathrm{1}}{\mathrm{9}}\:,\:{c}=\mathrm{0} \\ $$$$\Rightarrow\:\:\alpha\:=\:{h}−\frac{\frac{\mathrm{2}}{\mathrm{9}}\left(\frac{{h}}{\mathrm{9}}−{k}\right)}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{81}}} \\ $$$$\Rightarrow\:\:\boldsymbol{\alpha}=\boldsymbol{{h}}−\frac{\mathrm{9}}{\mathrm{41}}\left(\frac{\boldsymbol{{h}}}{\mathrm{9}}−\boldsymbol{{k}}\right) \\ $$$${and}\:\:\:\boldsymbol{\beta}\:=\:\boldsymbol{{k}}+\frac{\mathrm{81}}{\mathrm{41}}\left(\frac{\boldsymbol{{h}}}{\mathrm{9}}−\boldsymbol{{k}}\right) \\ $$$${eq}.\:{of}\:{line}\:{through}\:{BC}\:{is} \\ $$$${y}+{h}\:=\:\frac{\beta+{h}}{\alpha+{k}}\left({x}+{k}\right) \\ $$$${As}\:{this}\:{passes}\:{through}\:{P}\:\left({a},{b}\right) \\ $$$${b}+{h}\:=\:\frac{\beta+{h}}{\alpha+{k}}\left({a}+{k}\right) \\ $$$${replacing}\:\alpha,\:\beta\:{in}\:{terms}\:{of}\:{h},{k} \\ $$$${and}\:{changing}\:{h},{k}\:{to}\:{x},{y}\:{we}\:{get} \\ $$$${the}\:{desired}\:{locus}\:{of}\:{A}\left({third}\:{vertex}\right) \\ $$$$\boldsymbol{{b}}+\boldsymbol{{h}}=\frac{\boldsymbol{{k}}+\frac{\mathrm{81}}{\mathrm{41}}\left(\frac{\boldsymbol{{h}}}{\mathrm{9}}−\boldsymbol{{k}}\right)+\boldsymbol{{h}}}{\boldsymbol{{h}}−\frac{\mathrm{9}}{\mathrm{41}}\left(\frac{\boldsymbol{{h}}}{\mathrm{9}}−\boldsymbol{{k}}\right)+\boldsymbol{{k}}}\left(\boldsymbol{{a}}+\boldsymbol{{k}}\right) \\ $$$$\Rightarrow\:\left({b}+{x}\right)\left[{x}+{y}−\frac{\mathrm{9}}{\mathrm{41}}\left(\frac{{x}}{\mathrm{9}}−{y}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left({a}+{y}\right)\left[{x}+{y}+\frac{\mathrm{81}}{\mathrm{41}}\left(\frac{{x}}{\mathrm{9}}−{y}\right)\right] \\ $$$${locus}\:\:{found}\:{is} \\ $$$$\:\:\:\mathrm{40}\left(\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} \right)−\mathrm{41}\left(\boldsymbol{{b}}−\boldsymbol{{a}}\right)\left(\boldsymbol{{x}}+\boldsymbol{{y}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{9}\left(\boldsymbol{{ax}}−\boldsymbol{{by}}\right)=\mathrm{0}\:. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18
dear ajfour sir which app helps to draw ...for  example this problem...
$${dear}\:{ajfour}\:{sir}\:{which}\:{app}\:{helps}\:{to}\:{draw}\:…{for} \\ $$$${example}\:{this}\:{problem}… \\ $$
Commented by ajfour last updated on 06/Sep/18
lekh diagram.  i have understood how to use it  any confusion, u can ask me..
$${lekh}\:{diagram}. \\ $$$${i}\:{have}\:{understood}\:{how}\:{to}\:{use}\:{it} \\ $$$${any}\:{confusion},\:{u}\:{can}\:{ask}\:{me}.. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18
ok thank you...i have latent wish to see ourselves  we are virtually connected but never see each other  so if admin of this app give permission to upload  our picture atleast onece...i
$${ok}\:{thank}\:{you}…{i}\:{have}\:{latent}\:{wish}\:{to}\:{see}\:{ourselves} \\ $$$${we}\:{are}\:{virtually}\:{connected}\:{but}\:{never}\:{see}\:{each}\:{other} \\ $$$${so}\:{if}\:{admin}\:{of}\:{this}\:{app}\:{give}\:{permission}\:{to}\:{upload} \\ $$$${our}\:{picture}\:{atleast}\:{onece}…{i} \\ $$
Commented by MJS last updated on 06/Sep/18
we could upload a self portrait including a  math example...
$$\mathrm{we}\:\mathrm{could}\:\mathrm{upload}\:\mathrm{a}\:\mathrm{self}\:\mathrm{portrait}\:\mathrm{including}\:\mathrm{a} \\ $$$$\mathrm{math}\:\mathrm{example}… \\ $$
Commented by mondodotto@gmail.com last updated on 06/Sep/18
thanx
$$\mathrm{thanx}\: \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18
Commented by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18
Commented by behi83417@gmail.com last updated on 06/Sep/18
wonderful  idea!  waiting for sir: MrW3,ajfour,  sir MJS,prop.Abdo ,.....
$${wonderful}\:\:{idea}! \\ $$$${waiting}\:{for}\:{sir}:\:{MrW}\mathrm{3},{ajfour}, \\ $$$${sir}\:{MJS},{prop}.{Abdo}\:,….. \\ $$
Commented by MJS last updated on 06/Sep/18
Commented by MJS last updated on 06/Sep/18
the man is 1.95m tall, his beard is half as long  as Santa Claus′ . how many thorns has his  cactus if the temperature is 35°C?
$$\mathrm{the}\:\mathrm{man}\:\mathrm{is}\:\mathrm{1}.\mathrm{95m}\:\mathrm{tall},\:\mathrm{his}\:\mathrm{beard}\:\mathrm{is}\:\mathrm{half}\:\mathrm{as}\:\mathrm{long} \\ $$$$\mathrm{as}\:\mathrm{Santa}\:\mathrm{Claus}'\:.\:\mathrm{how}\:\mathrm{many}\:\mathrm{thorns}\:\mathrm{has}\:\mathrm{his} \\ $$$$\mathrm{cactus}\:\mathrm{if}\:\mathrm{the}\:\mathrm{temperature}\:\mathrm{is}\:\mathrm{35}°\mathrm{C}? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18
excellent...sir...i have thought a virtual image   of you sir...i shall post...what i imagined about  you sir...
$${excellent}…{sir}…{i}\:{have}\:{thought}\:{a}\:{virtual}\:{image}\: \\ $$$${of}\:{you}\:{sir}…{i}\:{shall}\:{post}…{what}\:{i}\:{imagined}\:{about} \\ $$$${you}\:{sir}… \\ $$$$ \\ $$
Commented by MJS last updated on 06/Sep/18
ha! I′d like to see it!
$$\mathrm{ha}!\:\mathrm{I}'\mathrm{d}\:\mathrm{like}\:\mathrm{to}\:\mathrm{see}\:\mathrm{it}! \\ $$
Commented by math khazana by abdo last updated on 07/Sep/18
we are waiting to know sir Tinkutara , sirAjfour ,sir Mrw ,sir  Raul ,sir Prakach jain ,sir Smai3...
$${we}\:{are}\:{waiting}\:{to}\:{know}\:{sir}\:{Tinkutara}\:,\:{sirAjfour}\:,{sir}\:{Mrw}\:,{sir} \\ $$$${Raul}\:,{sir}\:{Prakach}\:{jain}\:,{sir}\:{Smai}\mathrm{3}… \\ $$
Commented by malwaan last updated on 07/Sep/18
Commented by malwaan last updated on 07/Sep/18
that is me
$$\mathrm{that}\:\mathrm{is}\:\mathrm{me} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Sep/18
thank you sir...
$${thank}\:{you}\:{sir}… \\ $$

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