Consider the diagram shown below.
Equation of OC⇒y=m1x
Equation of OA⇒y=m2x
Given:
Equation of AC⇒lx+my=1
Equation of the sides ⇒ax2+2hxy+by2=0
Therefore,
m1+m2=−2hb
m1m2=ab
C is the intersection point of AC and OC. Therefore,
⇒lx+m(m1x)=1
⇒x=1l+mm1
⇒y=m1x=m1l+mm1
Also, A is the intersection point of AC and OA. Therefore,
⇒lx+m(m2x)=1
⇒x=1l+mm2
⇒y=m2x=m2l+mm2
Let P(α,β) be the mid-point of AC. Then,
α=12(1l+mm1+1l+mm2) ……. (1)
β=12(m1l+mm1+m2l+mm2) …… (2)
From equations (1) and (2), we have
⇒αβ=2l+m(m1+m2)m1(l+mm2)+m2(l+mm1)
⇒αβ=2l+m(−2hb)l(−2hb)+2m(ab)
⇒αβ=bl−mham−hl
We know that, diagonals of a parallelogram bisect each other. So, P will also lie on OB. Two points on OB are (0,0) and P(α,β). Therefore, equation of OB is,
⇒y=βαx
⇒y=am−hlbl−hmx
⇒y(bl−hm)=x(am−hl)
∴y(bl−hm)−x(am−hl)=0
Hence, this is the required answer.