**Each Pair Bisects the angle between the other pair** then $pq=?$

Question

If the pair of straight line $x^2-2pxy-y^2=0,x^2-2qxy-y^2=0$ are $\ni$ Each Pair Bisects the angle between the other pair then $pq=?$

I do not understand geometrically,mathematically the boldly written portion.

and it would be nice if some one give a solution,

I just only know one formula: angle between pair of st. line passing through origin.

Answer 1

$(0, 0)$ satisfies both equations so all these are lines through the origin. Further if $m_i$ represents the slopes of the first set of lines (which is $\frac{y}x$),
$$x^2-2pxy-y^2=0 \implies m_i^2+2pm_i-1 = 0 \implies m_1 m_2 = -1, m_1+m_2 = -2p$$

This means the lines are perpendicular. Similarly, you have if $n_1, n_2$ represent the slopes of the second set of lines, $n_1 + n_2 = -2q, n_1 n_2 = -1$, and again these lines are perpendicular.

So if the lines are bisecting each other's angles, the angles must be all of form $\alpha, \alpha + \frac{\pi}4, \alpha + \frac{\pi}2, \alpha + \frac{3\pi}4$. This gives $ m_i = \dfrac{1+n_i}{1-n_i}$. Using this and simplifying gives

$4pq = (m_1+m_2)(n_1+n_2) = -4 \implies pq = -1$

Answer 2

Geometry representation is

L(1a) of the first pair bisects the angle between L(2a) & L(2b) pair. Thus, u = v.

Similarly, L(2b) does the same thing to the L(1x) pair.