If the straight line through the point P (3, 4) makes an angle π/6 with the x-axis and meets the line 12x + 5y + 10 = 0 at Q, find the length PQ.

#### Solution

Here,

\[\left( x_1 , y_1 \right) = P \left( 3, 4 \right), \theta = \frac{\pi}{6} = {30}^\circ\]

So, the equation of the line is

\[\frac{x - x_1}{cos\theta} = \frac{y - y_1}{sin\theta}\]

\[ \Rightarrow \frac{x - 3}{\cos {30}^\circ} = \frac{y - 4}{\sin {30}^\circ}\]

\[ \Rightarrow \frac{x - 3}{\frac{\sqrt{3}}{2}} = \frac{y - 4}{\frac{1}{2}}\]

\[ \Rightarrow x - \sqrt{3}y + 4\sqrt{3} - 3 = 0\]

Let PQ = r

Then, the coordinates of Q are given by \[\frac{x - 3}{\cos30^\circ} = \frac{y - 4}{\sin30^\circ} = r\]

\[\Rightarrow x = 3 + \frac{\sqrt{3}r}{2}, y = 4 + \frac{r}{2}\]

Thus, the coordinates of Q are \[\left( 3 + \frac{\sqrt{3}r}{2}, 4 + \frac{r}{2} \right)\].

Clearly, the point Q lies on the line 12x + 5y + 10 = 0.

\[\therefore 12\left( 3 + \frac{\sqrt{3}r}{2} \right) + 5\left( 4 + \frac{r}{2} \right) + 10 = 0\]

\[ \Rightarrow 66 + \frac{12\sqrt{3} + 5}{2}r = 0\]

\[ \Rightarrow r = \frac{- 132}{5 + 12\sqrt{3}}\]

∴ PQ = \[\left| r \right|\] = \[\frac{132}{5 + 12\sqrt{3}}\]