# Straight lines

1. Sep 19, 2010

### zorro

1. The problem statement, all variables and given/known data

If the pair of lines x^2-2pxy-y^2=0 and x^2-2qxy-y^2=0 are such that each pair bisects the angle between the other pair, then what is the value of pq?

2. Relevant equations

3. The attempt at a solution

I don't understand what 'each pair bisects the angle between the other pair' mean.
Is it that each line of first pair bisects the angle between the two lines of other pair (obtuse and acute angle)?.
How do I proceed with this question?

2. Sep 19, 2010

### Mentallic

Yes pretty much. The lines in each equation are perpendicular to each other and this can be proven by showing that the first line satisfies:

$$y=(-p\pm\sqrt{p^2+1})x$$ and thus if $$m_1=-p+\sqrt{p^2+1}$$ and $$m_2=-p-\sqrt{p^2+1}$$ then $$m_1m_2=-1$$

And obviously since it's saying that p and q are chosen such that each pair of lines bisect each other, if we take any one line, then the other line from the other equation will have an angle difference of $\pi/4$ radians (since bisecting an angle of $\pi/2$ is half of that).

So we have $$m_p=-p\pm\sqrt{p^2+1}$$ and $$m_q=-q\pm\sqrt{q^2+1}$$.

since the gradient $$m_p=tan\theta$$ then $$m_q=tan(\theta+\frac{\pi}{4})=\frac{1+tan\theta}{1-tan\theta}=\frac{1+m_p}{1-m_p}$$

Can you proceed from here?

3. Sep 20, 2010

### zorro

In my book, it is done this way.

The second pair must be identical with (x^2-y^2)/[1-(-1)] = xy/-p

i.e. x^2 + (2/p)xy - y^2=0. Consequently, 2/p=-2q
i.e. pq=-1

I don't understand any thing from this. What is done here?
And what is the next step in your method?

4. Sep 21, 2010

### Mentallic

Neither can I

"The second pair must be identical with (x^2-y^2)/[1-(-1)] = xy/-p"

This line has completely stumped me on what they mean. Does it give any other information?

For my solution we have the equation

$$m_q=\frac{1+m_p}{1-m_p}$$ where $$m_q=-q\pm\sqrt{q^2+1}, m_p=-p\pm\sqrt{p^2+1}$$

Substituting mq into the equation:

$$-q\pm\sqrt{q^2+1}=\frac{1+m_p}{1-m_p}$$

Now solve for q, simplify, then substitute for mp and simplify further till you get the desired result.

5. Sep 22, 2010

### zorro

Unfortunately, nothing is mentioned.
Anyway I got the answer by your method (though it is a bit long :p)
Thanks alot.

6. Sep 22, 2010

### Mentallic

I agree, it is. I'd also like to know what your book meant by that... You gotta love how they don't even bother to give a reasonable explanation, while with other solutions they take you through it step by step so slowly, it makes you jump pages at a time to get to the point.

7. Sep 22, 2010

### zorro

Yeah, very true.

8. Sep 26, 2010

### zorro

You wanted to know what my book meant by that...
For ax^2 +2hxy + by^2 = 0 the angle bisector is given as...

x^2 - y^2 / ( a-b) = xy / h

This is a formula I got from my friend.

9. Sep 26, 2010

### Mentallic

And how was that formula derived?