Question

Find the locus of a point P that moves in such a way that the segment OP, where O is the origin, has slope $\sqrt 3 $.
A.$x - \sqrt 3 y = 0$
B. $x + \sqrt 3 y = 0$
C. $\sqrt 3 x + y = 0$
D. $\sqrt 3 x - y = 0$

Answer 1

Hint: Write the formula of a slope of a line, then equate that with $\sqrt 3 $. Then integrate the obtained equation and obtain the value of y with an integrating constant. Substitute 0 for x and 0 for y in the obtained equation to calculate the value of the integrating constant.

Formula Used:
The slope of a line is denoted by $\dfrac{{dy}}{{dx}}$ .
$\int {dx = x} $

Complete step by step solution:
The slope of a line is denoted by $\dfrac{{dy}}{{dx}}$.
Therefore,
$\dfrac{{dy}}{{dx}} = \sqrt 3 $
$dy = \sqrt 3 dx$
Integrate both sides of the equation,
$\int {dy} = \int {\sqrt 3 dx} $
$y = \sqrt 3 x + C$ , C is an integrating constant.
Substitute (0,0) for (x, y) in the obtained equation of y to obtain the value of C.
$0 = \sqrt 3 .0 + C$
Hence, $C = 0$ .
Therefore, the required result is $\sqrt 3 x - y = 0$.

Option ‘D’ is correct

Note: Another method is, write the slope intercept form of the line OP as $y = mx + c$ , where m is the slope and c is the y intercept. Given that $m = \sqrt 3 $ . So, the equation becomes $y = \sqrt 3 x + c$ , now the line passes through (0,0) so substitute (0,0) for (x, y) to obtain the value of c.
Hence, c=0. So, the equation will be $\sqrt 3 x - y = 0$.