Looking for a more efficient way to solve the Straight Line Equation problem.

Question

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: $\dfrac{132}{12\sqrt3+5}$

Answer 1

HINT:

$12x=-5(y+2)\iff\dfrac x{-5}=\dfrac{y+2}{12}=c$(say)

$\implies x=-5c,y=12c-2$ which is any point on $12x+5y+10=0$

Now the gradient of $PQ$ $$\tan\dfrac\pi6=\dfrac{12c-2-4}{-5c-3}$$

Find $c$ and hence $|PQ|$

Answer 2

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

$$ \vert PQ\vert=\dfrac{132}{12\sqrt3+5}=\frac{12}{37}(12\sqrt{3}-5) $$

Answer 3

Building on the idea in John Wayland Bales's answer, there is an easier way to get to the correct result. Here is John's picture:

We are given that the line $PQ$ is at an angle of $\pi/6$, or $30$ degrees. Remember that the 30-60-90 triangle is half an equilateral triangle, has side ratios $1:2:\sqrt{3}$. That means the line has a slope of $1/\sqrt{3}$. The slope of the other line is easy to see from its equation, and is $-12/5$. The $y=4$ line intersects that line at point $R$, for which the x-coordinate is then easy to calculate, making $R=(-2.5,4)$.

Let x be the horizontal distance between $P$ and $Q$. Using the fact that $|PR|=5.5$ and that if we go from $R$ to $Q$ to $P$ then the y coordinate does not change, we get the equation: $$-\frac{12}{5}(5.5-x) + \frac{1}{\sqrt{3}}x = 0$$ This gives $$x= \frac{66\sqrt{3} }{ (5 + 12\sqrt{3})}$$

Using the $30$ degree triangle ratios again, this means that the hypotenuse $PQ$ is $\frac{2}{\sqrt{3}}x$, or $$|PQ| = \frac{132 }{ (5 + 12\sqrt{3})}$$