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 ‌‌
