Question

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.

Answer 1

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

So, the equation of the line is

$$\begin{gathered} \frac{x-x_1}{\mathrm{cos\theta }}=\frac{y-y_1}{\mathrm{sin\theta }} \\ \Rightarrow \frac{x-3}{\cos {30}^{\circ }}=\frac{y-4}{\sin {30}^{\circ }} \\ \Rightarrow \frac{x-3}{{\displaystyle \frac{\sqrt{3}}{2}}}=\frac{y-4}{{\displaystyle \frac{1}{2}}} \\ \Rightarrow x-\sqrt{3}y+4\sqrt{3}-3=0 \end{gathered}$$

Let PQ = r
Then, the coordinates of Q are given by

$\frac{x-3}{\cos 30°}=\frac{y-4}{\sin 30°}=r$

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

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.

$$\begin{gathered} \therefore 12\left(3+\frac{\sqrt{3}r}{2}\right)+5\left(4+\frac{r}{2}\right)+10=0 \\ \Rightarrow 66+\frac{12\sqrt{3}+5}{2}r=0 \\ \Rightarrow r=\frac{-132}{5+12\sqrt{3}} \end{gathered}$$

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