Let's know the line L1: y = - 1 / a times x + 2. L2: y = - ax + 1 + a (a is not equal to 0). If L1 is perpendicular to L2, find the value of A

Let's know the line L1: y = - 1 / a times x + 2. L2: y = - ax + 1 + a (a is not equal to 0). If L1 is perpendicular to L2, find the value of A

The slope of line L1 = - 1 / A, the slope of line L2 = - A, if L1 is perpendicular to L2, then the slope product should be - 1, but (- 1 / a) * (- a) = 1, which is in contradiction with the theorem. Therefore, the value of a cannot be obtained

We know that L1: x + ay = 2A + 2 and L2: ax + y = a + 1
If L1 is parallel to L2, find the distance between the two parallel lines

Solution
If L1 / / L2
be
1-a²=0
∴a=1
Or a = - 1
When a = - 1
x-y=0
Coincidence with - x + y = 0
∴a=1
Namely
x+y-4=0
x+y-2=0
The distance between two parallel lines is
d=/-4+2//√2=√2

Given that the line L1: x + ay + 6 = 0 and the line L2: (A-2) x + 3Y + 2A = 0, then the sufficient and necessary condition of L1 ‖ L2 is that a equals ()
A. 3b. - 1C. - 1 or 3D. 1 or - 3

According to the meaning of the question, if L1 ‖ L2, then 1 × 3 = a × (A-2), the solution can be a = - 1 or 3, otherwise, when a = - 1, the line L1: X-Y + 6 = 0, its slope is 1, the line L2: - 3x + 3y-2 = 0, its slope is 1, and L1 and L2 do not coincide, then L1 ‖ L2, when a = 3, the line L1: x + 3Y + 6 = 0, the line L2: x +

Given the line L1: x + ay + 6 = 0 and L2: (A-2) x + 3Y + 2A = 0, then the distance between the two lines is

∵L1//L2
∴3/a=(a-2)/1
We get 3 = a-178; - 2A
a²-2a-3=0
(a-3)(a+1)=0
A = 3, rounding off
a=-1
L2:-3x+3y-2=0
x-y+2/3=0
L1：x-y+6=0
d=|2/3-6|√2=8√2/3
The distance between the two lines is 8 √ 2 / 3