Q.The equations of the lines passing through the point $(1,0)$ and at a distance $\dfrac{\sqrt{3}}{2}$ from the origin, are
(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 these.

To find the equations of lines, we first define a general line passing through the given point using its slope, then use the distance condition to determine the possible slopes.

The key idea is to combine the point-slope form of a line with the formula for the distance of a point from a line.

1. Let the slope of the line be $m$. Since the line passes through $(1,0)$, its equation is $y - 0 = m(x - 1)$, which simplifies to $mx - y - m = 0$.
2. The distance of this line from the origin $(0,0)$ is given as $\frac{\sqrt{3}}{2}$. Using the distance formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$, we have:

3. Squaring both sides to solve for $m$:

$$ 4m^2 = 3(m^2 + 1) \implies 4m^2 = 3m^2 + 3 \implies m^2 = 3 \implies m = \pm \sqrt{3} $$ …

The problem asks us to find the equations of lines that satisfy two conditions: they pass through a specific point $(1,0)$, and they are at a particular distance $\frac{\sqrt{3}}{2}$ from the origin $(0,0)$.

The core idea is to first represent all possible lines passing through the given point. This creates a "family" of lines, each distinguished by its slope. Then, we use the second condition – the distance from the origin – to find the specific slopes that satisfy the requirement.

1. Represent the family of lines:

   A line passing through a point $(x_1, y_1)$ can be generally written in the point-slope form:

   $y - y_1 = m(x - x_1)$

   Here, the given point is $(1,0)$, so we substitute $x_1=1$ and $y_1=0$:

   $y - 0 = m(x - 1)$

   $y = m(x - 1)$

2. Convert to the general form of a line:

   To use the distance formula from a point to a line, we need the line's equation in the general form $Ax + By + C = 0$.

   Rearranging $y = m(x - 1)$:

   $y = mx - m$

   $mx - y - m = 0$

   This is our general line equation, where $A=m$, $B=-1$, and $C=-m$.

3. Apply the distance formula:

   The distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by:

   $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

   In this problem, the point is the origin $(0,0)$, so $x_0=0$ and $y_0=0$. The given distance is $d = \frac{\sqrt{3}}{2}$.

   Substituting these values, along with $A=m$, $B=-1$, $C=-m$, into the distance formula:

   $\frac{\sqrt{3}}{2} = \frac{|m(0) + (-1)(0) + (-m)|}{\sqrt{m^2 + (-1)^2}}$

   $\frac{\sqrt{3}}{2} = \frac{|-m|}{\sqrt{m^2 + 1}}$

4. Solve for the slope $m$:

   We need to solve the equation $\frac{|-m|}{\sqrt{m^2 + 1}} = \frac{\sqrt{3}}{2}$ for $m$.

   Since $|-m| = |m|$, we have:

   $\frac{|m|}{\sqrt{m^2 + 1}} = \frac{\sqrt{3}}{2}$

   To eliminate the absolute value and the square root, we square both sides of the equation:

   $\left(\frac{|m|}{\sqrt{m^2 + 1}}\right)^2 = \left(\frac{\sqrt{3}}{2}\right)^2$

   $\frac{m^2}{m^2 + 1} = \frac{3}{4}$

   Now, cross-multiply:

   $4m^2 = 3(m^2 + 1)$

   $4m^2 = 3m^2 + 3$

   $4m^2 - 3m^2 = 3$

   $m^2 = 3$

   Taking the square root of both sides gives two possible values for $m$:

   $m = \pm\sqrt{3}$ …