Question

A straight line parallel to the line y = √3 x passes through Q(2,3) and cuts the line 2x + 4y - 27 = 0 at P. Then the length of the line segment PQ is

• A. 2√3 + 1
• B. √3 +1
• C. 2√3 -1
• D. √3 - 1

Answer 1

The Correct Option is C

To find the distance between point $Q(2, 3)$ and point $P$, the intersection of the line through $Q$ parallel to $y = \sqrt{3}x$ and the line $2x + 4y - 27 = 0$, we proceed as follows:

1. Equation of the Line Through $Q$:
The line parallel to $y = \sqrt{3}x$ has the form $y = \sqrt{3}x + c$. Since it passes through $Q(2, 3)$, substitute $x = 2$, $y = 3$:

$ 3 = \sqrt{3} \cdot 2 + c $
$ c = 3 - 2\sqrt{3} $
Thus, the equation is:

$ y = \sqrt{3}x + 3 - 2\sqrt{3} $
Rewrite in standard form:

$ \sqrt{3}x - y + 3 - 2\sqrt{3} = 0 $

2. Finding the Intersection Point $P$:
Solve the system of equations for the intersection of $\sqrt{3}x - y + 3 - 2\sqrt{3} = 0$ and $2x + 4y - 27 = 0$:

$ \sqrt{3}x - y = 2\sqrt{3} - 3 \quad (1) $
$ 2x + 4y = 27 \quad (2) $
From (1), express $y$:

$ y = \sqrt{3}x - 2\sqrt{3} + 3 $
Substitute into (2):

$ 2x + 4(\sqrt{3}x - 2\sqrt{3} + 3) = 27 $
$ 2x + 4\sqrt{3}x - 8\sqrt{3} + 12 = 27 $
$ (2 + 4\sqrt{3})x = 27 + 8\sqrt{3} - 12 = 15 + 8\sqrt{3} $
$ x = \frac{15 + 8\sqrt{3}}{2 + 4\sqrt{3}} $
Rationalize:

$ x = \frac{15 + 8\sqrt{3}}{2(1 + 2\sqrt{3})} \cdot \frac{1 - 2\sqrt{3}}{1 - 2\sqrt{3}} = \frac{(15 + 8\sqrt{3})(1 - 2\sqrt{3})}{2(1 - 4 \cdot 3)} = \frac{15 - 30\sqrt{3} + 8\sqrt{3} - 16 \cdot 3}{-22} $
$ = \frac{15 - 48 - 22\sqrt{3}}{-22} = \frac{-33 - 22\sqrt{3}}{-22} = \frac{33 + 22\sqrt{3}}{22} = \frac{3 + 2\sqrt{3}}{2} $
Substitute $x$ into (1) to find $y$:

$ y = \sqrt{3} \left( \frac{3 + 2\sqrt{3}}{2} \right) + 3 - 2\sqrt{3} = \frac{3\sqrt{3} + 6}{2} + 3 - 2\sqrt{3} $
$ = \frac{3\sqrt{3} + 6 + 6 - 4\sqrt{3}}{2} = \frac{12 - \sqrt{3}}{2} $
Thus, point $P$ is:

$ P = \left( \frac{3 + 2\sqrt{3}}{2}, \frac{12 - \sqrt{3}}{2} \right) $

3. Calculating the Distance $PQ$:
The distance $PQ$ is:

$ PQ = \sqrt{\left( \frac{3 + 2\sqrt{3}}{2} - 2 \right)^2 + \left( \frac{12 - \sqrt{3}}{2} - 3 \right)^2} $
Simplify the coordinates:

$ x_P - x_Q = \frac{3 + 2\sqrt{3} - 4}{2} = \frac{-1 + 2\sqrt{3}}{2} $
$ y_P - y_Q = \frac{12 - \sqrt{3} - 6}{2} = \frac{6 - \sqrt{3}}{2} $
So:

$ PQ = \sqrt{\left( \frac{-1 + 2\sqrt{3}}{2} \right)^2 + \left( \frac{6 - \sqrt{3}}{2} \right)^2} $
$ = \sqrt{\frac{(-1 + 2\sqrt{3})^2 + (6 - \sqrt{3})^2}{4}} $
$ = \sqrt{\frac{(1 - 4\sqrt{3} + 12) + (36 - 12\sqrt{3} + 3)}{4}} $
$ = \sqrt{\frac{1 + 12 + 36 + 3 - 4\sqrt{3} - 12\sqrt{3}}{4}} = \sqrt{\frac{52 - 16\sqrt{3}}{4}} = \sqrt{13 - 4\sqrt{3}} $

4. Verifying the Distance:
$ (2\sqrt{3} - 1)^2 = (2\sqrt{3})^2 - 2 \cdot 2\sqrt{3} \cdot 1 + 1^2 = 12 - 4\sqrt{3} + 1 = 13 - 4\sqrt{3} $
Since $\sqrt{13 - 4\sqrt{3}} = 2\sqrt{3} - 1$, 

Final Answer:
The distance $PQ$ is $2\sqrt{3} - 1$.