Given the two equations L1: MX + 3Y = 0, L2 "MX + (m-2) y + = 0", when L1 is parallel to and perpendicular to L2, the range of M is calculated

Given the two equations L1: MX + 3Y = 0, L2 "MX + (m-2) y + = 0", when L1 is parallel to and perpendicular to L2, the range of M is calculated

When m = 2, the two lines are
2X + 3Y = 0, x = 0, neither parallel nor vertical, round off
When m! = 2
L1 slope K1 = - M / 3, L2 slope K2 = m / (2-m)
When two lines are parallel
K1 = K2, the solution is m = 0 or 5, because when m = 0, the two lines are y = 0, so it is rounding off
So m = 5
When two lines are perpendicular
k1*k2=-1
M & sup2; + 3m-6 = 0
Solution
Substituting into the solution, M = (- 3 ± root 33) / 2