# [SOLVED] Straight lines(coordinate geometry)?

• Sep 10th 2008, 06:15 AM
fardeen_gen
[SOLVED] Straight lines(coordinate geometry)?
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.
• Sep 10th 2008, 08:48 AM
Laurent
The equation of the line $\displaystyle \mathcal{L}:$$\displaystyle ax+by=c$ can be written $\displaystyle \overrightarrow{OM}\cdot\vec{u}=c$, where $\displaystyle \vec{u}={a\choose b}$ (and $\displaystyle M$ is the point we are describing). Then $\displaystyle \vec{u}$ is a vector orthogonal to the line (if $\displaystyle M,N\in\mathcal{L}$, then $\displaystyle \overrightarrow{MN}\cdot\vec{u}=\overrightarrow{ON }\cdot\vec{u}-\overrightarrow{OM}\cdot\vec{u}=c-c=0$), hence the projection of $\displaystyle O$ on $\displaystyle \mathcal{L}$ is the point $\displaystyle P$ such that $\displaystyle \overrightarrow{OP}=\lambda\vec{u}$ and $\displaystyle P\in\mathcal{L}$, i.e. $\displaystyle \lambda\vec{u}\cdot\vec{u}=c$, so $\displaystyle \lambda=\frac{c}{|\vec{u}|^2}$ and $\displaystyle \overrightarrow{OP}=\frac{c}{|\vec{u}|^2}\vec{u}$, and the distance between the origin and the line $\displaystyle \mathcal{L}$ is $\displaystyle d=OP=|\overrightarrow{OP}|=\frac{|c|}{|\vec{u}|}$. 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 $\displaystyle x\cos\theta+y\sin\theta=k\sin\theta\cos\theta$ (multiplying both sides by $\displaystyle \sin\theta\cos\theta$), so that $\displaystyle |\vec{u}|=1$ for both lines, so you have to check that $\displaystyle (k\cos2\theta)^2+4(k\sin\theta\cos\theta)^2=k^2$. I guess you know how to do this.

Laurent.
• Sep 10th 2008, 08:59 AM
Plato
Here is a second way.
Using the standard distance formula we get:
$\displaystyle p = \frac{{\left| { - k\cos \left( {2\phi } \right)} \right|}}{{\sqrt {\sin ^2 \left( \phi \right) + \cos ^2 \left( \phi \right)} }} = \left| k \right|\left| {\cos \left( {2\phi } \right)} \right| \Rightarrow \quad p^2 = k^2 \cos ^2 \left( {2\phi } \right)$
and $\displaystyle q = \frac{{\left| { - k} \right|}}{{\sqrt {\sec ^2 \left( \phi \right) + cse^2 \left( \phi \right)} }} = \left| k \right|\sqrt {\cos ^2 \left( \phi \right)\sin ^2 \left( \phi \right)}$.
From that it follows $\displaystyle q^2 = k^2 \cos ^2 \left( {\phi } \right)\sin ^2 \left( {\phi } \right) \Rightarrow \quad 4q^2 = k^2 \sin ^2 \left( {2\phi } \right)$.
Now you finish.
• Sep 10th 2008, 10:16 AM
Laurent
Quote:

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)
• Sep 10th 2008, 10:49 AM
Plato
Quote:

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.