Let the area of the triangle formed by (1,2),(−3,0) and any point on the line x−2y+k=0 is 5 sq. units. If possible values of k are k1,k2 then locus of foot of the perpendicular drawn from (k1,0) to a variable line passing through (0,k2) is

Let the area of the triangle formed by (1,2),(−3,0) and any point on

1.	Let the area of the triangle formed by (1,2),(−3,0) and any point on the line x−2y+k=0 is 5 sq. units. If possible values of k are k1,k2 then locus of foot of the perpendicular drawn from (k1,0) to a variable line passing through (0,k2) is
Answer» Let the area of the triangle formed by $(1, 2), (- 3, 0)$ and any point on the line $x - 2 y + k = 0$ is $5$ sq. units. If possible values of $k$ are $k_{1}, k_{2}$ then locus of foot of the perpendicular drawn from $(k_{1}, 0)$ to a variable line passing through $(0, k_{2})$ is