The locus of a point such that the sum of its distances from two given perpendicular lines is equal to $2$ units in the first quadrant is

  • A
    $x + y + 2 = 0$
  • B
    $x + y = 2$
  • C
    $x - y = 2$
  • D
    None of these

Explore More

Similar Questions

Let $A=(4,0)$ and $B=(0,12)$ be two points in the plane. The locus of a point $C(x, y)$ such that the area of $\triangle ABC$ is $18$ sq units is

If the sum of the perpendicular distances of a variable point $P(x, y)$ from the lines $x + y - 5 = 0$ and $3x - 2y + 7 = 0$ is always $10$,show that $P$ must move on a line.

Difficult
View Solution

If $A=(2,3)$ and $B=(-4,5)$ are two fixed points,then the locus of a point $P$ such that the area of $\triangle PAB$ is $12$ square units is

$A$ variable line passes through a fixed point $(a, b)$ and meets the coordinate axes at $A$ and $B$. The locus of the point of intersection of lines drawn through $A$ and $B$ parallel to the coordinate axes is:

Difficult
View Solution

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

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