Straight line and equation (13 18:7:26) The line L1: x + y + a = 0, L2: x + ay + 1 = 0 and L3: ax + y + 1 = 0 can form a triangle, and the value range of a can be obtained

Straight line and equation (13 18:7:26) The line L1: x + y + a = 0, L2: x + ay + 1 = 0 and L3: ax + y + 1 = 0 can form a triangle, and the value range of a can be obtained

The line L1: x + y + a = 0, L2: x + ay + 1 = 0, L3: ax + y + 1 = 0 can form a triangle
L1 slope K1 = - 1, intercept B1 = - A
The slope of L2 K2 = - 1 / A, intercept B2 = - 1 / A
The slope of L3 K3 = - A, intercept B3 = - 1
k1≠k2≠k3,a≠±1,
The intersection of L1 and L2 (- 1-A, 1) is not on L3
a(-1-a)+1+1≠0,a≠-2,a≠1
In conclusion, the value range of a is a ≠± 1, a ≠ - 2

Straight line and equation (13 18:4:25)
If a (- 4,2), B (3,1), find the coordinates of point C, and judge the shape of △ ABC

If y = 2x is the line where the bisector of ∠ C in △ ABC lies, then a is a & acute; (X & acute;, Y & acute;) on BC, let a be a & acute; (X & acute;, Y & acute;) then (Y & acute; - 2) / (X & acute; + 4) = - 1 / 2 (1) (Y & acute; + 2) / 2 = 2 (X & AC