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If x-y axes are inclined at angle $\omega$ then show that equation $$x^2 + 2xy\cos \omega + y^2 \cos 2\omega = 0$$ represents a pair of perpendicular straight lines passing through origin.

Let two lines with inclination to axis x as $m, n$.

Then the condition of perpendicularity of these lines is $mn + 1 + (m + n)cos \omega = 0$

After some manipulation we get,

$$m = -{n\cos\omega + 1\over \cos \omega + n}$$

So the lines are $$y - nx\ \text{and} \ y + \left({n\cos\omega + 1\over \cos \omega + n}\right)x $$

So pair of straight lines would be

$$(y - nx) \times \left(y + \left({n\cos\omega + 1\over \cos \omega + n}\right)x\right) = 0$$

On Expanding,

$$y^2(\cos \omega - n) - xyn(\cos \omega + n) + xy(n\cos \omega + 1) + x^2(-n)(n\cos\omega + 1) = 0$$

On simplification,

$$y(\cos \omega - n) + xy(1 - n^2) + x^2(-n^2\cos \omega - n) = 0$$

How do I remove $n$ from this equation and get the "to be proven" equation ?

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    @victoria The coordinate system isn’t orthogonal (“$x$-$y$ axes are inclined at an angle $\omega$”) so $u\cdot v=0$ doesn’t mean that the vectors are orthogonal. A different inner product is called for. – 2017-02-20
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    @victoria The axes are oblique not rectangular. – 2017-02-20
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    @amd Can you give me some hint on this question, I have been struggling with these since I had learnt them. – 2017-02-20

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You’ve stated a condition for an arbitrary pair of lines to be orthogonal in the given coordinate system, so it’s not terribly surprising that there’s a left-over free variable. In this problem, though, you’re dealing with a specific pair of lines, so you need to compute their slopes and use those values for $m$ and $n$. The equation can be factored into $$(x+y\cos\omega)^2-(y\sin\omega)^2 = (x+y(\cos\omega+\sin\omega))(x+y(\cos\omega-\sin\omega))=0$$ from which the slopes of the two lines are $-(\cos\omega+\sin\omega)^{-1}$ and $-(\cos\omega-\sin\omega)^{-1}$. Substituting these values into your orthogonality condition eventually yields $0$ as required.

That said, working with slopes might not be the best approach since doing so omits the case of a line parallel to the $y$-axis, which you’ll then have to treat separately for a complete proof. I would solve this problem by transforming the equation to an orthogonal coordinate system. The mapping from a standard Cartesian coordinate system to the one in the problem is $$\begin{align}x&=x'-y'\cot\omega\\y&=y'\csc\omega.\end{align}$$ Substituting this into the equation of the lines and simplifying yields $x'^2-y'^2=0$, which can be seen to be a pair of perpendicular lines by factoring this equation.

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    How did you get those mapping ? – 2017-02-20
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    I inverted the mapping going in the other direction, which is easy to derive after staring at a drawing of the oblique coordinate axes for a bit: $x'=x+y\cos\omega$ and $y'=y\sin\omega$. – 2017-02-20
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    Oh thanks for the answer and comment. – 2017-02-20