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Find the locus of the circum-centre of a triangle whose two sides are along the coordinate axes and the third side passes through the point of intersection of the lines ax+by+c=0 and lx+my+n=0

Find the locus of the circum-centre of a triangle whose two sides are along the coordinate axes and the third side passes through the point of intersection of the lines ax+by+c=0 and lx+my+n=0

Grade:12

2 Answers

Vikas TU
14149 Points
3 years ago
As two sides are along co−ordinate tomahawks. Henceforth triangle OAB will 
be correct calculated. 
Leave locus alone circumcentre be (h,k) and condition of hypotenuse be xp+yq=1. 
As triangle is correct calculated henceforth circumcentre will be at the mid purpose of the hypotenuse. 
⇒(h,k)=(p+0/2,0+q/2) 
⇒p=2h and q=2k. 
⇒x/p+y/q=1 moves toward becoming x/2h+y/2k=1 
x/h+y/k=2 (i) 
Given that hypotenuse goes through ax+by+c=0 ii 
lx+my+n=0 iii 
(equation(ii)×l −equation(iii)×a, we get,alx+bly+cl−alx−amy−an=0 
⇒y(bl−am)=an−cl⇒y=an−cl/bl−am 
equation(ii)×m −equation(iii)×b, we get,amx+bmy+cm−blx−bmy−bn=0 
x(am−bl)=bn−cm 
⇒x=bn−cmam−bl 
Putting the estimation of (x,y) in equation(i), we get,1h(bn−cmam−bl)+1k(an−clbl−am)=2 
⇒k(bn−cm)(bl−am)+h(an−cl)(am−bl)hk(am−bl)(bl−am)=2 
⇒k(bn−cm)(bl−am)+h(an−cl)(am−bl)=2hk(am−bl)(bl−am) 
Replace h→x and k→y to get the condition of locus of circumcenter. 
⇒y(bn−cm)(bl−am)+x(an−cl)(am−bl)=2xy(am−bl)(bl−am) 
Nikhil Swami
26 Points
3 years ago
LET THE EQUATION OF THIRD LINE BE (ax+by+c)+ $(lx+my+n)=0 where $ is a parameterIt meets the x-axid at A where y=0 and x=(-$n+c)/$l+a

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