It is known that the slopes of two straight lines L1 and L2 are the two roots of the equation 3x ^ 2 + MX-3 = 0, m ∈ R Then L1, L2

It is known that the slopes of two straight lines L1 and L2 are the two roots of the equation 3x ^ 2 + MX-3 = 0, m ∈ R Then L1, L2

Let the slopes of L1 and L2 be K1 and K2
According to Weida's theorem
k1 * k2 = -3/3 = -1
So L1 and L2 are perpendicular to each other

Let a (- 1,0) and B (1,0) pass through two points a and B respectively, and the product of the slopes of L1 and L2 is - 4?
I don't know how to do this kind of problem very well. If I can add 50 points to the basic idea of doing this kind of problem (which way to think about it)

Intersection P (x, y), then L1 slope (y-0) / (x + 1) L2 slope (y-0) / (x-1) so [y / (x + 1)] = - 4Y & sup2; = 4 (x-1) (x + 1) = 4x & sup2; - 4x & sup2; - Y & sup2; / 4 = 1, obviously the slope cannot be 0, otherwise the product is 0, so y is not equal to 0, so x & sup2; - Y & sup2; / 4 = 1, excluding (- 1,0), (1