Help neede in coordinate geometry

[metalInferno]

Help neede in coordinate geometry!

Homework Statement

here is the question :

If the pairs of lines $$x^2-2pxy-y^2$$ and $$x^2-2qxy-y^2$$ are such that each pair bisects the angle between the other pair , then pq equals ...??the answer somehow is -1. i think it uses the concept of homogenisation (if u ppl know what it is . searched the net but found not a single page on it ) . if u know this concept please explain it 2 me and how u come to this answer .
thanx in advance

Homework Equations

The Attempt at a Solution

 

[Shooting Star]

It’s difficult to give a brief hint for this. Draw a neat diagram.

Since angular bisectors of a pair of lines are always perpendicular to each other, it’s obvious that both pairs are mutually perpendicular pairs of lines. Name one of the pairs L1, L2 and the other one L3, L4. Let L3 be between L1 and L2 when we are going from L1 to L2 in the CCW direction. So, if we go CCW, it will be L1, L3, L2, L4.

It’s also obvious that the angles between adjacent lines are all 45 deg, i.e., angle between L1 and L3 is 45 deg etc.

1.
If m1 and m2 are the slopes of L1 and L2, then their combined eqn is (y - m1x)(y - m2x) = 0 =>
y^2 – (m1 + m2)xy – m1m2x^2 =0.
Comparing this with the first of the eqns given, we get that, 2p = -(m1 + m2).
Similarly, 2q = -(m3 + m4).

2.
Let’s also remember that if the angle between L1 and L3 is positive 45 deg in the CCW sense, then tan 45 = (m3 – m1)/(1 + m1m3) => m1m3 = m3 – m1 -1. There will be four eqns like this.

3.
2p*2q = (m1 + m2)(m3 + m4). Expand this and you’ll get four terms like m2m3. Write that as m3 – m1 -1. All the terms will cancel out except (-1)*4 => pq = -1.

(Why are math questions being posted in the Physics page? Is it because the posters find the Physics page more interesting?)

 

[Shooting Star]

Faster mehtod

I think this solution will appeal more to you. Actually, I just remembered the formulae for these things. This one’s fast and probably the one you should use during tests. (And it uses homogenization too…)

The eqn of the angular bisectors of the pair of lines ax^2 + 2hxy + by^2 = 0 is given by the eqn
(x^2 – y^2)/(a – b) = xy/h.

Comparing the two given eqns, and knowing that each set are the bisectors of the others, the answer is immediately arrived at.