If the lines joining origin and point of intersection of curves ax2+2hxy+by2+2gx=0 and a1x2+2h1xy+b1y2+2g1x=0 are mutually perpendicular,then prove that-g(a1+b1)=g1(a+b)

Hello student,
the given curves are ax²+2hxy+by²+2gx=0...............(1)
anda1x²+2h1xy+b1y²+2g1x=0...............(2)
Eliminating 1^stdegree terms from (1) and (2)
multiplying (1) by g¹and (2) by g and subtract them we get
g¹(ax²+2hxy+by²+2gx)-g(a1x²+2h1xy+b1y²+2g1x)=0
(ag¹-a1g)x²+2(hg¹-h¹g)xy+(bg¹-b¹g)y²=0
These lines are mutually perpendicular if coeff of x²+coeff of y²=0
Hence we get g(a1+b1)=g1(a+b)