The point moves such that the area of the triangle formed by it with the points $(1, 5)$ and $(3, -7)$ is $21$ sq. unit. The locus of the point is

The co-ordinates of the vertices $A$ and $B$ of an isosceles triangle $ABC (AC = BC)$ are $(-2,3)$ and $(2,0)$ respectively. $A$ line parallel to $AB$ and having a $y$ -intercept equal  to $\frac{43}{12}$ passes through $C$, then the co-ordinates of $C$ are :-

The opposite angular points of a square are $(3,\;4)$ and $(1,\; - \;1)$. Then the co-ordinates of other two points are

Let $B$ and $C$ be the two points on the line $y+x=0$ such that $B$ and $C$ are symmetric with respect to the origin. Suppose $A$ is a point on $y -2 x =2$ such that $\triangle ABC$ is an equilateral triangle. Then, the area of the $\triangle ABC$ is

The area of a parallelogram formed by the lines $ax \pm by \pm c = 0$, is

A point moves so that square of its distance from the point $(3, -2)$ is numerically equal to its distance from the line $5x - 12y = 13$. The equation of the locus of the point is