$A$ point on the straight line $3x + 5y = 15$ which is equidistant from the coordinate axes will lie only in

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 -

If the algebraic sum of the perpendicular distances from the points $(2,0)$,$(0,2)$,and $(1,1)$ to a variable line is zero,then the variable line always passes through a fixed point. The coordinates of that point are

Given points $A(6,0)$,$B(0,4)$,and $O$ as the origin,find the locus of a point $P(x, y)$ such that the area of $\triangle POB$ is $2$ times the area of $\triangle POA$.

$A$ variable straight-line $L$ with negative slope passes through the point $(4,9)$ and cuts the positive coordinate axes in $A$ and $B$. If $O$ is the origin,then the minimum value of $OA+OB$ is

Let $P(2, -3)$ and $Q(-2, 1)$ be the vertices of the $\Delta PQR$. If the centroid of $\Delta PQR$ lies on the line $2x + 3y = 1$,then the locus of $R$ is