Find the equation to the pair of angle bisectors of the pair of lines $(ax+by)^2=3(bx-ay)^2$.
Efforts: $$(ax+by)^2=3(bx-ay)^2$$ After simplifying, I got: $$x^2(a^2-3b^2)+8abxy+y^2(b^2-3a^2)=0$$ Now, what should I do next?
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You can simplify and try the hard way. Here's what I will do: Here: $$(ax + by)^2 - 3*(bx − ay)^2 = 0$$ this is of the form $$a^2-b^2 = (a+b)(a-b)$$ => the two lines are $$(ax + by + sqrt(3)*bx -sqrt(3)*ay = 0)$$ and $$(ax + by -sqrt(3)*bx + sqrt(3)*ay = 0)$$
Iaamuser user
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Win Vineeth
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That will go well. Except for some particular cases (where some denominators are $0$): Let the slopes be $\tan C$ and $\tan D$. We compute $\tan (C+D)=2 a b/(a^2-b^2)$ ,whence $\tan (C+D)/2\in {b/a,-a/b}.$ – DanielWainfleet Feb 07 '16 at 06:21
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Taking the square root on both sides, we obtain $$ax+by=-\sqrt3(bx-ay) \quad \text{and} \quad ax+by=\sqrt3(bx-ay)$$ and the two lines are $$ (a+\sqrt3b)x + (b-\sqrt3 a)y = 0\quad\text{and}\quad (a-\sqrt3b)x + (b+\sqrt3 a)y = 0. $$ The equation for the bisectors is known to be $$ \frac{|(a+\sqrt3b)x + (b-\sqrt3 a)y|}{\sqrt{(a+\sqrt3b)^2+(b-\sqrt3 a)^2}}=\frac{|(a-\sqrt3b)x + (b+\sqrt3 a)y|}{\sqrt{(a-\sqrt3b)^2+(b+\sqrt3 a)^2}}, $$ which in fact is equivalent to $$ |(a+\sqrt3b)x + (b-\sqrt3 a)y|=|(a-\sqrt3b)x + (b+\sqrt3 a)y|. $$ This gives $$ bx -ay=0 \quad\text{or}\quad ax + by=0. $$
John B
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I only got $bx-ay=0$, how did you get $ax+by=0$, could you please elaborate? – Iaamuser user Feb 09 '16 at 16:45
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Note that $(a+\sqrt3b)x + (b-\sqrt3 a)y=\pm[(a-\sqrt3b)x + (b+\sqrt3 a)y]$. – John B Feb 09 '16 at 16:50
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The modulus comes from the original equation. But then you need to solve it, like $|x|=|y|$, which gives $x=y$ or $x=-y$. Just a compact way of writing. – John B Feb 09 '16 at 16:56