Given equation, 2x2+kxy−6y2+3x+y+1=0 is a pair of straight line. ∴ |
2
k
2
3
2
k
2
−6
1
2
3
2
1
2
1
|=0 ⇒ |
4
k
3
k
−12
1
3
1
2
|=0 ⇒4(−24−1)−k(2k−3)+3(k+36)=0 ⇒−100−2k2+3k+3k+108=0 ⇒2k2−6k−8=0 ⇒k2−3k−4=0 ⇒(k−4)(k+1)=0,∴k=4{∵k>0} ∴ Equation of line is ⇒2x2+4xy−6y2+3x+y+1=0 ⇒(2x−2y+1)(x+3y+1)=0 ⇒2x−2y+1=0 and x+3y+1=0 Solving equation, we get x=