The perpendicular bisector of the line segment joining the points $P(1, 4)$ and $Q(k, 3)$ has a $y$-intercept of $-4$. Then a possible value of $k$ among the following is:

$A$ straight line meets the $X$ and $Y$ axes at the points $A$ and $B$ respectively. If $AB = 6$ units,then the locus of the point $P$ which divides the line segment $AB$ such that $AP : PB = 2 : 1$ is

$P(x, y)$ moves such that the area of the triangle formed by $P, Q(a, 2a)$ and $R(-a, -2a)$ is equal to the area of the triangle formed by $P, S(a, 2a)$ and $T(2a, 3a)$. The locus of $P$ is a straight line given by:

$O$ is the origin and $A$ is the point $(3,4)$. If a point $P(x,y)$ moves such that the line segment $OP$ is always parallel to the line segment $OA$,then the equation of the locus of $P$ is:

Show that the path of a moving point such that its distances from two lines $3x - 2y = 5$ and $3x + 2y = 5$ are equal is a straight line.

If the line $\frac{x}{a} + \frac{y}{b} = 1$ is a variable line such that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$,then the locus of the foot of the perpendicular from the origin to the line is: