A point equidistant from the points (2, 0) an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the point be $P(x, y)$. The distance from $P$ to $(2, 0)$ is equal to the distance from $P$ to $(0, 2)$.Using the distance formula: $\sqrt{(x-

Vedclass · Accepted Answer

$(2, 2)$

Vedclass · Answer

(D) Let the point be $P(x, y)$. The distance from $P$ to $(2, 0)$ is equal to the distance from $P$ to $(0, 2)$.Using the distance formula: $\sqrt{(x-2)^2 + (y-0)^2} = \sqrt{(x-0)^2 + (y-2)^2}$.Squaring both sides: $(x-2)^2 + y^2 = x^2 + (y-2)^2$.Expanding the terms: $x^2 - 4x + 4 + y^2 = x^2 + y^2 - 4y + 4$.Simplifying: $-4x = -4y$,which implies $x = y$.Checking the options,the point $(2, 2)$ satisfies $x = y$ and is equidistant from $(2, 0)$ and $(0, 2)$.

$A$ point equidistant from the points $(2, 0)$ and $(0, 2)$ is

Similar Questions

If a point $P$ is equidistant from the points $A(a + b, b - a)$ and $B(a - b, a + b)$,find the locus of $P$.

If $2a + b + 3c = 0$,then the line $ax + by + c = 0$ passes through the fixed point that is

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:

$A$ point moves in the $xy$-plane such that the sum of its distances from two mutually perpendicular lines is always equal to $5$ units. The area (in sq units) enclosed by the locus of the point is

If $A=(5,3)$,$B=(3,-2)$ and a point $P$ is such that the area of the triangle $PAB$ is $9$,then the locus of $P$ represents

Vedclass Test Series

Exam Paper Generator

Online Exam Module