[SOLVED] Straight lines(coordinate geometry)?

**fardeen_gen** · September 10th 2008, 06:15 AM

If p and q are the lengths of the perpendiculars from the origin to the lines x cos θ - y sin θ = k cos 2θ and x sec θ + y cosec θ = k , respectively, prove that p^2 + 4q^2 = k^2.

**Laurent** · September 10th 2008, 08:48 AM

The equation of the line $\mathcal{L}:$ $ax+by=c$ can be written $\overrightarrow{OM}\cdot\vec{u}=c$ , where $\vec{u}={a\choose b}$ (and $M$ is the point we are describing). Then $\vec{u}$ is a vector orthogonal to the line (if $M,N\in\mathcal{L}$ , then $\overrightarrow{MN}\cdot\vec{u}=\overrightarrow{ON }\cdot\vec{u}-\overrightarrow{OM}\cdot\vec{u}=c-c=0$ ), hence the projection of $O$ on $\mathcal{L}$ is the point $P$ such that $\overrightarrow{OP}=\lambda\vec{u}$ and $P\in\mathcal{L}$ , i.e. $\lambda\vec{u}\cdot\vec{u}=c$ , so $\lambda=\frac{c}{|\vec{u}|^2}$ and $\overrightarrow{OP}=\frac{c}{|\vec{u}|^2}\vec{u}$ , and the distance between the origin and the line $\mathcal{L}$ is $d=OP=|\overrightarrow{OP}|=\frac{|c|}{|\vec{u}|}$ . This is a useful result to remember.

Using this fact, the exercise is just the verification of a simple trigonometric formula. It is even easier if you rewrite the second equation as $x\cos\theta+y\sin\theta=k\sin\theta\cos\theta$ (multiplying both sides by $\sin\theta\cos\theta$ ), so that $|\vec{u}|=1$ for both lines, so you have to check that $(k\cos2\theta)^2+4(k\sin\theta\cos\theta)^2=k^2$ . I guess you know how to do this.

Laurent.

**Plato** · September 10th 2008, 08:59 AM

Here is a second way.
Using the standard distance formula we get:
$p = \frac{{\left| { - k\cos \left( {2\phi } \right)} \right|}}{{\sqrt {\sin ^2 \left( \phi \right) + \cos ^2 \left( \phi \right)} }} = \left| k \right|\left| {\cos \left( {2\phi } \right)} \right| \Rightarrow \quad p^2 = k^2 \cos ^2 \left( {2\phi } \right)$
and $q = \frac{{\left| { - k} \right|}}{{\sqrt {\sec ^2 \left( \phi \right) + cse^2 \left( \phi \right)} }} = \left| k \right|\sqrt {\cos ^2 \left( \phi \right)\sin ^2 \left( \phi \right)}$ .
From that it follows $q^2 = k^2 \cos ^2 \left( {\phi } \right)\sin ^2 \left( {\phi } \right) \Rightarrow \quad 4q^2 = k^2 \sin ^2 \left( {2\phi } \right)$ .
Now you finish.

**Laurent** · September 10th 2008, 10:16 AM

Originally Posted by Plato

Here is a second way.

(actually our ways are pretty close, because my first paragraph only proves the usual distance formula, which I presumed fardeen_gen did not know)

**Plato** · September 10th 2008, 10:49 AM

Originally Posted by Laurent

(actually our ways are pretty close, because my first paragraph only proves the usual distance formula, which I presumed fardeen_gen did not know)

I think that it is more likely that fardeen_gen does not know vectors to that extent.

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