<s>[INST] Looking for a more efficient way to solve the Straight Line Equation problem. 
If the straight line through the point $P(3,4)$ makes an angle $\dfrac{\pi}{6}$ with the $x$ axis and meets the line $12x +5y + 10=0$ at Q, find the length of PQ.

My method: 
Equation of the given line is;
$y- \sqrt3x+3\sqrt3-4=0$ // using Point slope form. 
Now, solving the simultaneous equations gives us the point of intersection $R$. And using the Distance formula, I can find the length $PR$. However, I feel this problem can be solved in a more simpler manner. What would be an efficient approach?  
PS: The answer is: $\dfrac{132}{12\sqrt3+5}$
 [/INST] The given line contains the point $P=(3,4)$ and has slope $m=\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$ so its equation is
$$ -\sqrt{3}x+3y-12+3\sqrt{3}=0\tag{1} $$
The second line has equation
$$ 12x+5y+10=0\tag{2} $$
The two equations intersect at a point $Q$ and we wish to find the distance from $P$ to $Q$. 
The coordinates of $Q$ are quite messy. Sketching the graph might suggest an alternate way to approach the problem. 

Notice that the horizontal line $y=4$ appears to cross the second line at the point $R=(-2.5,4)$ which we quickly verify by substitution into the second equation. Now we have a triangle $\triangle PQR$.
We know that $\angle QPR=\dfrac{\pi}{6}$ and that $PR=5.5$. If we can find $\angle QRP$ we can use the Law of Sines to find $PQ$.
The angle  $\angle QRP$ is the same as the angle between the normal vectors of equation $(2)$ and equation $0\cdot x+1\cdot y=4$ which can be found using the vector version of the Law of Cosines.
The normal vector of $(2)$ is $\mathbf{u}=(12,5)$ and the normal vector of $0\cdot x+1\cdot y=4$ is $\mathbf{v}=(0,1)$. So we have
$$ \cos(\angle QRP)=\frac{(12,5)\cdot(0,1)}{\Vert(12,5)\Vert\,\Vert(0,1)\vert}=\frac{5}{13} $$
Therefore,
$$ \sin(\angle QRP)=\frac{12}{13} $$
Now we know that
$$\angle PQR=\pi-\left(\frac{\pi}{6}+\arcsin\left(\frac{12}{13}\right)\right)$$
So
\begin{eqnarray}
\sin\left(\angle PQR\right)
&=&\sin\left(\frac{5\pi}{6}\right)\cdot\frac{5}{13}-\cos\left(\frac{5\pi}{6}\right)\cdot\frac{12}{13}\\
&=&\frac{12\sqrt{3}+5}{26}
\end{eqnarray}
By the Law of Sines we have
$$ \frac{\vert PQ\vert}{12/13}=\frac{11/2}{\left(\frac{12\sqrt{13}+5}{26}\right)} $$
Working through the arithmetic gives
$$ \vert PQ\vert=\dfrac{132}{12\sqrt3+5}=\frac{12}{37}(12\sqrt{3}-5) $$
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