Straight Lines and Pair of Straight Lines

Given equation is 2x2+xy−6y2=0
‌⇒‌‌(2x−3y)(x+2y)=0
‌⇒‌‌2x−3y=0‌ and ‌x+2y=0
Now, ax2−7xy−3y2=0 and 2x2+xy−6y2 has exactly one line in common.
So, either it has 2x−3y=0 or x+2y=0
Case 1ax2−7xy−3y2=0
‌⇒‌‌(2x−3y)(px+qy)=0
‌⇒‌‌2px2+(2q−3p)xy−3qy2=0
On comparing, we get
‌2p=a,2q−3p=−7,−3q=−3
⇒‌‌q=1
So, 2(1)−3p=−7
‌⇒‌‌3p=9
‌⇒‌‌p=3
and ‌‌2p=a
⇒‌‌a=2(3)=6
Case II x+2y=0
So, ax2−7xy−3y2=0
‌⇒‌‌(x+2y)(rx+sy)=0
‌⇒‌‌rx2+(s+2r)xy+2sy2=0
On comparing, we get
r‌=a,s+2r=−7,2s=−3
⇒‌‌s‌=‌

−3

2

So, s=+2r=−7
⇒2r‌=−7−s
‌=−7+‌

3

2

=−‌

11

2

⇒r‌=‌

−11

4

So, a=r=‌

−11

4

∴a=6,p=3,q=1
So, the lines are (2x−3y)(3x+y)=0
⇒‌‌6x2−7xy−3y2=0
Now, the bisectors are given by
‌‌

x2−y2

6−(−3)

=‌

xy

−‌ 7 2

⇒‌

x2−y2

9

=‌

2xy

−7

⇒−7x2+7y2=18xy
⇒7x2+18xy−7y2=0