The locus of the mid-points of the perpendiculars drawn from points on the line $x=2y$ to the line $x=y$ is:

  • A
    $2x-3y=0$
  • B
    $7x-5y=0$
  • C
    $5x-7y=0$
  • D
    $3x-2y=0$

Explore More

Similar Questions

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:

$A \equiv (\cos \theta, \sin \theta)$ and $B \equiv (\sin \theta, -\cos \theta)$ are two points. The locus of the centroid of $\triangle OAB$,where $O$ is the origin,is

$A$ variable line passing through $(l, m)$ intersects the coordinate axes at the points $A$ and $B$. If the line drawn parallel to $Y$-axis through $A$ and parallel to $X$-axis through $B$ meet at $P$,then the locus of $P$ is

The locus of a point which is at a distance of $2$ units from the line $2x - 3y + 4 = 0$ and at a distance of $\sqrt{13}$ units from the point $(5, 0)$ is:

If $P_1, P_2, P_3, \ldots, P_n$ are $n$ points on the line $y=x$ all lying in the first quadrant,such that $OP_n = n(OP_{n-1})$ ($O$ is origin),$OP_1 = 1$ and $P_n = (2520 \sqrt{2}, 2520 \sqrt{2})$,then $n=$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo