If a point $P$ on the line $3x + 5y = 15$ is equidistant from the coordinate axes,then $P$ lies

The locus of the point $P$ which is equidistant from $3x + 4y + 5 = 0$ and $9x + 12y + 7 = 0$ is:

The locus of a point such that the sum of its distances from two given perpendicular lines is equal to $2$ units in the first quadrant is

$A$ line segment joining a point $A$ on $x$-axis to a point $B$ on $y$-axis is such that $AB=15$. If $P$ is a point on $AB$ such that $\frac{AP}{PB}=\frac{2}{3}$,then the locus of $P$ is:

If the locus of the centroid of the triangle with vertices $A(a, 0)$,$B(a \cos t, a \sin t)$ and $C(b \sin t, -b \cos t)$ ($t$ is a parameter) is $9x^2 + 9y^2 - 6x = 49$,then the area of the triangle formed by the line $\frac{x}{a} + \frac{y}{b} = 1$ with the coordinate axes is

The line $\frac{x}{a} + \frac{y}{b} = 1$ moves in such a way that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{2c^2},$ where $a, b, c \in R_0$ and $c$ is a constant. Then,the locus of the foot of the perpendicular from the origin on the given line is -