Find the equation of a line passing through the point $(4, 3)$,which cuts a triangle of minimum area from the first quadrant.

$A$ straight line passes through a fixed point $(h, k)$. The locus of the foot of the perpendicular drawn from the origin to this line is:

The equation of the locus of the foot of the perpendicular drawn from the origin to a line passing through a fixed point $(a, b)$ is:

Let $A$ be the point $(0,4)$ and $B$ be a moving point on the $x$-axis. Let $M$ be the midpoint of $AB$ and let the perpendicular bisector of $AB$ meet the $y$-axis at $R$. The locus of the midpoint $P$ of $MR$ is

$A$ straight line $L$ with negative slope passes through the point $(1,1)$ and cuts the positive coordinate axes at the points $A$ and $B$. If $O$ is the origin,then the minimum value of $OA + OB$ as $L$ varies,is

Every point $(x, y)$ on the curve $3x + 2y - 3xy = 0$ is the centroid of a triangle formed by the coordinate axes and a line $(L)$ intersecting both the coordinate axes. Then all such lines $(L)$