The locus of a variable point which forms a triangle of fixed area with two fixed points is

Let $A$ be the point $(0,4)$ in the $xy$-plane and let $B$ be the point $(2t, 0)$. Let $L$ be the midpoint of $AB$ and let the perpendicular bisector of $AB$ meet the $y$-axis at $M$. Let $N$ be the midpoint of $LM$. Then the locus of $N$ 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

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

If the equation of the locus of a point equidistant from the points $(a_1, b_1)$ and $(a_2, b_2)$ is $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$,then the value of $c$ is