Q.

If (α, β) is the centriod of the triangle formed by the lines ax2 + 2hxy + by2 = 0 and lx + my = 1, prove that
αblhm=βamhl=23bl22hlm+am2

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Detailed Solution

Let ax2+2hxy+by2=0 represents the lines
l1x+m1y=0..(1) l2x+m2y=0.(2)l1x+m1yl2x+m2y=ax2+2hxy+by2
Comparing like terms on  both sides we get
l1l2=a,m1m2=b,l1m2+l2m1=2h
Given line is lx + my = 1
Clearly the origin O is the point of intersection of (1) and (2)
Let A be the point of intersection of (1) and (3)
By the method of cross multiplication
          x   y            1                m10l1m1 m1lm 
xm10=y0+l1=1l1mlm1x=m1l1mlm1,y=l1l1mlm1A=m1l1mlm1,l1l1mlm1
Similarly the point of intersection of (2) and (3) is
B=m2l2mlm2,l2l2mlm2
Centroid of OAB is G = (α,β) =
m1l1mlm1m2l2mlm23,l1l1mlm1+l2l2mlm23α=13m1l1mlm1m2l2mlm2,β=13l1l1mlm1+l2l2mlm2α=13m1l2mlm2m2l1mlm1l1mlm1l2mlm2=132m1m2ll2m1ml1m2ml1l2m2l1m2+l2m1lm+m1m2l2
=2bl2hm3bl22hlm+am2αblhm=23bl22hlm+am2
Similarly βamhl=23bl22hlm+am2
αblhm=βamhl=23bl22hlm+am2

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