(D) Let the point $P$ be $(x, y)$.Since $O$ is the origin $(0, 0)$,the slope of the segment $OP$ is given by $m = \frac{y - 0}{x - 0} = \frac{y}{x}$.G

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(D) Let the point $P$ be $(x, y)$.Since $O$ is the origin $(0, 0)$,the slope of the segment $OP$ is given by $m = \frac{y - 0}{x - 0} = \frac{y}{x}$.Given that the slope is $\sqrt{3}$,we have $\frac{y}{x} = \sqrt{3}$.Multiplying both sides by $x$,we get $y = \sqrt{3}x$.Rearranging the terms,we obtain $\sqrt{3}x - y = 0$.Thus,the required locus is $\sqrt{3}x - y = 0$.

Class 11 Mathematics Straight Line Locus of Point

Easy

The locus of a point $P(x, y)$ which moves in such a way that the segment $OP$,where $O$ is the origin $(0, 0)$,has a slope of $\sqrt{3}$ is:

A
$x - \sqrt{3}y = 0$
B
$x + \sqrt{3}y = 0$
C
$\sqrt{3}x + y = 0$
D
$\sqrt{3}x - y = 0$

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Straight Line

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Locus of Point

The point $P$ is equidistant from $A(1, 3)$,$B(-3, 5)$,and $C(5, -1)$. Then $PA$ is equal to:

DifficultAP EAMCET 2003

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DifficultJEE MAIN 2019

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If the equation of the locus of a point equidistant from $(a_1, b_1)$ and $(a_2, b_2)$ is $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$,find the value of $c$.

Medium

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$ABC$ is a variable triangle such that $A$ is $(1, 2)$,and $B$ and $C$ lie on the line $y = x + \lambda$ (where $\lambda$ is a variable). The locus of the orthocenter of triangle $ABC$ is:

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If the locus of a point which is equidistant from the coordinate axes forms a triangle with the line $y=3$,then the area of the triangle is

MediumAP EAMCET 2025

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The locus of a point $P(x, y)$ which moves in such a way that the segment $OP$,where $O$ is the origin $(0, 0)$,has a slope of $\sqrt{3}$ is:

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