Straight Lines and Pair of Straight Lines Part 1

Given, A(0,6) and B(2t,0)
Mid-point of AB=M(t,3)
Equation of perpendicular bisector of AB passes through M.
∴y−3=

t

3

(x−t) ...(i)
So, C(0,3−

t2

3

)
Intersection of Eq. (i) on Y - axis
C(0,3−

t2

3

)
Let mid-point of MC is (h, k).
Then,(h,k)=(

t

2

,3−

t2

6

)
⇒h=

t

2

,k=3−

t2

6

Eliminating t, we get
2h2=3(3−k)
Locus of (h,k)
2x2=3(3−y)
⇒2x2+3y−9=0