IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

sonalisapal

wht is the distance between orthocentre nd incentre of the triangle formed by the lines x+y=2 xy=zero

AJAYAVSS

is it 2/root(2)+1,the whole root(2)+1 is in denominator

krishna.gopal

X=0, y= 0 and x+y =2 make a right angle triangle of sides 1,1, root(2) with vertecies (1,0),(0,1),(0,0).
The orthocenter of right angled triangle is right angled vertex only = (0,0)
x cordinate of incenter is given by (ax1+bx2+cx3)/(a+b+c)=(1/(2+root(2)))
y cordinate of incenter is given by (ay1+by2+cy3)/(a+b+c)=(1/(2+root(2)))

Hence distance is root(2)/(2+root(2))=1/(root(2)+1)

dik1101

Your analogy is correct.
But The answer acc. to me is incorrect.
The vertices should be (2,0), (0,2), (0,0)
So, the answer becomes 2(root(2))/2+(root(2))

Asmita

Krishna.gopal sir has only committed a calculation mistake while calculating the coordinates of the incentre.

Incentre = 2/(root2+2),2/(root2+2)

As orthocentre is (0,0)

therefore by dis. formula

the answer comes out to be 2/(root2+1) OR 2root2/(2+root2)

krishna.gopal

Well sorry for the mistake. It is actually (2,0),(0,2) and (0,0). So answer will be twice what i told earlier. And my name is krishna gopal. That dot between krishna and gopal comes only in my login id, not in my real name

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