Let $A=(4,0)$ and $B=(0,12)$ be two points in the plane. The locus of a point $C(x, y)$ such that the area of $\triangle ABC$ is $18$ sq units is

$P(x, y)$ moves such that the area of the triangle formed by $P, Q(a, 2a)$ and $R(-a, -2a)$ is equal to the area of the triangle formed by $P, S(a, 2a)$ and $T(2a, 3a)$. The locus of $P$ is a straight line given by:

Show that the path of a moving point such that its distances from two lines $3x - 2y = 5$ and $3x + 2y = 5$ are equal is a straight line.

If $A=(0,1), B=(1,2), C=(-2,1)$,then the equation of the locus of a point $P(x,y)$ such that the area of triangle $PAB$ equals the area of triangle $PAC$ is:

$A$ variable line passes through a fixed point $(a, b)$ and meets the coordinate axes at $A$ and $B$. The locus of the point of intersection of lines drawn through $A$ and $B$ parallel to the coordinate axes is:

The locus of the incentre of the triangle formed by the lines $xy-4x-4y+16=0$ and $x+y=5$ is