# 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. - Mathematics

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 = $\left| r \right|$ = $\frac{132}{5 + 12\sqrt{3}}$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 23 The straight lines
Exercise 23.8 | Q 2 | Page 65