Question

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

Answer 1

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

$\text{So, the equation of the line is}$

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

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

$\Rightarrow \frac{x-3}{\frac{\sqrt{3}}{2}}=\frac{y-3}{\frac{1}{2}}$

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

$LetPQ=r$

$\text{Then, the coordinates of Q are given by}$

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

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

$\text{Thus, the coordinates of Q are}$

$(3+\frac{\sqrt{3}r}{2},4+\frac{r}{2})$

$\text{Clearly, the point Q lies on the line}12x+5y+10=0$

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

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

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

$\therefore PQ=|r|=\frac{132}{5+12\sqrt{3}}$