Given that, equation of pair of straight lines is (ax+by)2−3(bx−ay)2=0 ⇒a2x2+b2y2+2abxy −3(b2x2+a2y2−2abxy)=0 ⇒(a2−3b2)x2−(b2−3a2)y2+8abxy=0 Let the equation of line be y=m1x,y=m2x. ∴m1+m2=
−4ab
b2−3a2
[
m1+m2=−
2h
b
m1m2=
a
b
] and m1m2=
a2−3b2
b2−3a2
Now, (m1−m2)2=(m1+m2)2−4m1m2 =
16a2b2
(b2−3a2)2
−
4(a2−3b2)
(b2−3a2)
=
4
(b2−3a2)2
[4a2b2−(a2−3b2)(b2−3a2)] =
4
(b2−3a2)2
[4a2b2−(a2b2−3a4−3b4+9a2b2)] =
4
(b2−3a2)2
[−6a2b2+3a4+3b4] =
24
(b2−3a2)2
(a2−b2)2 ∴(m1−m2)=
2√3
(b2−3a2)
⋅(a2−b2) The intersecting point on the lines y=m1x, y=m2x and ax+by+c=0 are A(0,0),B(