Question

A straight line through the point P(1,2) makes an angle $\theta$ with the positive X-axis in anti-clockwise direction and meets the line $x+\sqrt{3}y-2\sqrt{3}=0$ at Q. If $PQ = \frac{1}{2}$, then $\theta=$

• A. $\frac{\pi}{6}$
• B. $\frac{5\pi}{6}$
• C. $\frac{2\pi}{3}$
• D. $\frac{\pi}{3}$

Hint:

In problems involving distance along a line in parametric form, remember that the distance $d$ corresponds to a directed distance $r = \pm d$. You may need to check both cases, as one might lead to a valid solution within the given options while the other might not.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
This problem uses the parametric (or symmetric) form of a straight line. A point on a line at a certain distance from a given point can be represented using the angle the line makes with the x-axis.
Step 2: Key Formula or Approach
The parametric equation of a line passing through $(x_1, y_1)$ and making an angle $\theta$ with the positive x-axis is: \[ \frac{x-x_1}{\cos\theta} = \frac{y-y_1}{\sin\theta} = r \] where $|r|$ is the distance of the point $(x,y)$ from $(x_1, y_1)$. Any point on this line can be written as $(x_1 + r\cos\theta, y_1 + r\sin\theta)$. We are given $P(x_1, y_1) = (1,2)$ and the distance $PQ = |r| = 1/2$. This means $r$ could be $1/2$ or $-1/2$. We substitute the coordinates of point Q into the equation of the second line and solve for $\theta$.
Step 3: Detailed Explanation
The coordinates of point Q, which is at a directed distance $r$ from P(1,2), are given by: \[ x_Q = 1 + r\cos\theta, \quad y_Q = 2 + r\sin\theta \] The distance $PQ$ is given as $1/2$. We consider two cases for the directed distance $r$: $r=1/2$ and $r=-1/2$. Case 1: $r = 1/2$ The coordinates of Q are $(1 + \frac{1}{2}\cos\theta, 2 + \frac{1}{2}\sin\theta)$. Since Q lies on the line $x+\sqrt{3}y-2\sqrt{3}=0$, we substitute these coordinates into the equation: \[ (1 + \frac{1}{2}\cos\theta) + \sqrt{3}(2 + \frac{1}{2}\sin\theta) - 2\sqrt{3} = 0 \] \[ 1 + \frac{1}{2}\cos\theta + 2\sqrt{3} + \frac{\sqrt{3}}{2}\sin\theta - 2\sqrt{3} = 0 \] \[ 1 + \frac{1}{2}\cos\theta + \frac{\sqrt{3}}{2}\sin\theta = 0 \] \[ \cos\theta + \sqrt{3}\sin\theta = -2 \] This equation is of the form $a\cos\theta+b\sin\theta=c$. The maximum value of the left side is $\sqrt{1^2+(\sqrt{3})^2} = 2$. The minimum value is $-2$. The expression $\cos\theta + \sqrt{3}\sin\theta$ equals its minimum value of $-2$. This occurs when $\cos\theta = -1/2$ and $\sin\theta = -\sqrt{3}/2$. The angle $\theta$ for this is $4\pi/3$. This is not among the options. Case 2: $r = -1/2$ The coordinates of Q are $(1 - \frac{1}{2}\cos\theta, 2 - \frac{1}{2}\sin\theta)$. Substitute these coordinates into the line's equation: \[ (1 - \frac{1}{2}\cos\theta) + \sqrt{3}(2 - \frac{1}{2}\sin\theta) - 2\sqrt{3} = 0 \] \[ 1 - \frac{1}{2}\cos\theta + 2\sqrt{3} - \frac{\sqrt{3}}{2}\sin\theta - 2\sqrt{3} = 0 \] \[ 1 - \frac{1}{2}\cos\theta - \frac{\sqrt{3}}{2}\sin\theta = 0 \] \[ 1 = \frac{1}{2}\cos\theta + \frac{\sqrt{3}}{2}\sin\theta \] \[ \cos\theta + \sqrt{3}\sin\theta = 2 \] The expression $\cos\theta + \sqrt{3}\sin\theta$ equals its maximum value of $2$. This can be written as $2(\frac{1}{2}\cos\theta + \frac{\sqrt{3}}{2}\sin\theta) = 2$, or $2\cos(\theta - \frac{\pi}{3})=2$. \[ \cos(\theta - \frac{\pi}{3}) = 1 \] This implies that $\theta - \frac{\pi}{3} = 2n\pi$ for some integer $n$. \[ \theta = 2n\pi + \frac{\pi}{3} \] For $n=0$, we get $\theta = \frac{\pi}{3}$. This is one of the options. Step 4: Final Answer
By considering the parametric form of the line and the two possible directed distances $r=\pm 1/2$, we find that $r=-1/2$ leads to the equation $\cos\theta + \sqrt{3}\sin\theta = 2$, which has the solution $\theta=\pi/3$.