Through the point $(4, 5)$,a straight line is drawn making positive intercepts on the coordinate axes. The area of the triangle thus formed is least,when the ratio of the intercepts on the $X$ and $Y$ axes is

$A(-4,0)$ and $B(4,0)$ are two fixed points. $C$ and $D$ are two points on the $Y$-axis such that $CD=4$ and $C$ is a point below $D$. Then the locus of the point of intersection of the lines $AC$ and $BD$ is

$A$ moving line intersects the lines $x+y=0$ and $x-y=0$ at the points $A$ and $B$ respectively, such that the area of the triangle with vertices $(0,0)$, $A$, and $B$ has a constant area $C$. The locus of the mid-point of $AB$ is given by the equation:

The locus of a point which moves so that it is always equidistant from the points $A(a, 0)$ and $B(-a, 0)$ is

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

Let $BC$ be a fixed line segment in the plane. The locus of a point $A$ such that the $\triangle ABC$ is isosceles,is (with finitely many possible exceptional points)