Equation of given pair of straight lines S=3x2−2kxy+y2=0 and the line L=2x−y−6=0 On solving, we get 3x2−2kx(2x−6)+(2x−6)2=0 ⇒‌‌(7−4k)x2+12(k−2)x+36=0 Let the intersection points are A(x1,y1) and B(x2,y2) ∴ The area of △ABC is
‌
1
2
|
0
0
0
x1
y1
1
x2
y2
1
|=36⇒|‌
y1
x1
−‌
y2
x2
|=‌
72
|x2x1|
∵‌‌m1+m2=2k,m1m2=3 and ‌
y1
x1
=m1 ‌
y2
x2
=m2 and |x1x2|=‌
36
|7−4k|
So, |m1−m2|=‌
72
36
|7−4k|
⇒‌‌(m1+m2)2−4m1m2=4(7−4k)2
(on squaring both sides) ⇒‌‌4k2−12=4(49+16k2−56k)
⇒15k2−56k+52=0⇒15k2−30k−26k+52=0
⇒15k(k−2)−26(k−2)=0 ⇒‌‌k=2 (∵k is an integer)