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,

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

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

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

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

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

$$\therefore 12(3+\frac{\sqrt{3}r}{2})+5(4+\frac{r}{2})+10=0$$

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

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

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