If the sum of the distances of a point from two perpendicular lines in a plane is $1$,then its locus is:

If a point $P$ is equidistant from the points $A(a + b, b - a)$ and $B(a - b, a + b)$,find the locus of $P$.

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

$A$ point equidistant from the points $(2, 0)$ and $(0, 2)$ is

$A$ line cuts the $x$-axis at $A(7, 0)$ and the $y$-axis at $B(0, -5)$. $A$ variable line $PQ$ is drawn perpendicular to $AB$ cutting the $x$-axis at $P(a, 0)$ and the $y$-axis at $Q(0, b)$. If $AQ$ and $BP$ intersect at $R(h, k)$, the locus of $R$ is

$A$ variable line passes through a fixed point $(x_{1}, y_{1})$ and meets the axes at $A$ and $B$. If the rectangle $OAPB$ is completed, the locus of $P$ is, ($O$ being the origin of the system of axes).