The line L1: y = (K / 3-2k) × (x + 2) passing through the point (- 1,1) with a slope of 3 / K + 2 is parallel 2 and the line L: mx-m & # 178; y = 1, which is perpendicular to the point P (2,1), the equation is?

1.y=[k/(3-2k)](x+2)=[k/(3-2k)]x+2[k/(3-2k)]
∵ y = MX + B, where m is the slope; and the line is parallel to L1
∴k/(3-2k)=3/(k+2),
9-6k=k²+2k,k²+8k-9=2,(k+9)(k-1)=0,k=1or-9
2. By introducing P (2,1) into mx-m & # 178; y = 1, 2m-m & # 178; = 1, M & # 178; - 2m + 1 = 0, (m-1) &# 178; = 0, M = 1
The line L is X-Y = 1, y = X-1, slope = 1
∵ line L is perpendicular to line L 'and its slopes are reciprocal
The slope of line L 'is - 1
Straight line L ': (Y-1) = - 1 (X-2), y = - x + 3

Given the slope of the line L1, the line L2 passes through a (3a, - 2), B (0, a + 1), and L1 ⊥ L2, find the value of the real number a,

Let L1 slope be K1 and L2 slope be K2. Since L1 is perpendicular to L2, K * K2 = - 1, that is K2 = - 1 / K1
Let L2 equation y = k2x + B, substitute a (3a, - 2) B (0, a + 1) into it, and the solution is a = 3 * K1 / (3-k1)

It is known that the slopes of lines L1 and L2 are respectively a / (A-1), (1-A) / (2a + 3). (1) when a is what value, L1 is perpendicular to L2
(2) Is there a real number a so that L1 is parallel to L2? If so, find a. if not, explain the reason

L1 is perpendicular to L2
a/(a-1)*(1-a)/(2a+3).=-1
The solution is a = - 3 or 1
In the special case a = 1, one slope is 0 and the other does not exist
(2) Suppose there is a real number a such that L1 is parallel to L2, then
A / (A-1) = (1-A) / (2a + 3), i.e
3A ^ 2 + A + 1 = 0
Discriminant

Given that the line L1 passing through a (- 2,0) and B (1,3a) is perpendicular to the line L2 passing through P (0, - 1) and Q (a, - 2A), the value of real number a is obtained

L1 slope = (3a-0) / (1 + 2) = a
L2 slope = (- 2A + 1) / (A-0) = (- 2A + 1) / A
Vertical slope is negative reciprocal
So a * (- 2A + 1) / a = - 1
a=1

The line L1: y = (K / 3-2k) × (x + 2) passing through the point (- 1,1) with a slope of 3 / K + 2 is parallel 2 and the line L: mx-m & # 178; y = 1, which is perpendicular to the point P (2,1), the equation is?