TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

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
∵‌‌m1+m2=2k,m1m2=3 and ‌

y1

x1

=m1
‌

y2

x2

=m2 and |x1x2|=‌

36

|7−4k|

So, |m1−m2|=‌

72

36

|7−4k|
(on squaring both sides)
⇒‌‌4k2−12=4(49+16k2−56k)
⇒15k(k−2)−26(k−2)=0
⇒‌‌k=2 (∵k is an integer)