Question

The equations of the line passing through the point (1, 0) and at a distance $\dfrac{{\sqrt 3 }}{2}$ from the origin is:
A) $\sqrt 3 x + y - \sqrt 3 = 0,\sqrt 3 x - y - \sqrt 3 = 0$
B) $\sqrt 3 x + y + \sqrt 3 = 0,\sqrt 3 x - y + \sqrt 3 = 0$
C) $x + \sqrt 3 y - \sqrt 3 = 0,x - \sqrt 3 y - \sqrt 3 = 0$
D) None of the above

Answer 1

Hint: We will first write the general equation and then put in the point (1, 0) which is given to us. After that, we will just find distance in terms of m and put it equal to the given distance and thus we have the answer.

Complete step-by-step solution:
We know that the equation of a line passing through the point $\left( {{x_1},{y_1}} \right)$ is given by:-
$ \Rightarrow y - {y_1} = m\left( {x - {x_1}} \right)$, where m is the slope of the line.
We have the point (1, 0) and if we compare it to $\left( {{x_1},{y_1}} \right)$, then we get ${x_1} = 1$ and ${y_1} = 0$.
So, putting these values in the above equation, we will get:-
$ \Rightarrow y - 0 = m\left( {x - 1} \right)$
Simplifying the equation to get the following equation:-
$ \Rightarrow y - mx + m = 0$
$ \Rightarrow mx + y - m = 0$ ………………(1)
We know that distance of the line ax + by + c = 0 from the origin is given by: $d = \dfrac{{|c|}}{{\sqrt {{a^2} + {b^2}} }}$.
Therefore, as we compare the both of equation, we have a = m, b = 1 and c = -m.
So, the distance from origin will be: $d = \dfrac{{| - m|}}{{\sqrt {{m^2} + {1^2}} }}$
Since, we are already given that the distance of line from origin is $\dfrac{{\sqrt 3 }}{2}$ units.
$\therefore d = \dfrac{{| - m|}}{{\sqrt {{m^2} + {1^2}} }} = \dfrac{{\sqrt 3 }}{2}$
$ \Rightarrow \dfrac{{| - m|}}{{\sqrt {{m^2} + {1^2}} }} = \dfrac{{\sqrt 3 }}{2}$
Squaring both sides, we will get:-
$ \Rightarrow {\left( {\dfrac{{| - m|}}{{\sqrt {{m^2} + {1^2}} }}} \right)^2} = {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2}$
$ \Rightarrow \dfrac{{{m^2}}}{{{m^2} + 1}} = \dfrac{3}{4}$
Cross – multiplying the terms to get the following expression:-
$ \Rightarrow 4{m^2} = 3\left( {{m^2} + 1} \right)$
Simplifying the RHS by opening up the parenthesis on the RHS, we will then obtain:-
$ \Rightarrow 4{m^2} = 3{m^2} + 3$
$ \Rightarrow 4{m^2} - 3{m^2} = 3$
$ \Rightarrow {m^2} = 3$
$ \Rightarrow m = \pm \sqrt 3 $
Putting this in (1), we will get:-
$ \Rightarrow \sqrt 3 x + y - \sqrt 3 = 0$ or $ - \sqrt 3 x + y + \sqrt 3 = 0$
$ \Rightarrow \sqrt 3 x + y - \sqrt 3 = 0$ or $\sqrt 3 x - y - \sqrt 3 = 0$

$\therefore $ The correct option is (A).

Note: The students must remember the following formulas:-
Distance of the line ax + by + c = 0 from the origin is given by: $d = \dfrac{{|c|}}{{\sqrt {{a^2} + {b^2}} }}$.
The equation of a line passing through the point $\left( {{x_1},{y_1}} \right)$ is given by:-
$ \Rightarrow y - {y_1} = m\left( {x - {x_1}} \right)$, where m is the slope of the line.
The students might make the mistake of taking the value of m different in the same equation. Therefore, they must remember that they should put in both values one by one in the general equation to get the two lines.
Let us understand why there are two lines. It is because we have two possibilities that line may go from the above or below the origin.