Given the line L1: x + Y-1 = 0, L2: 2x-y + 3 = 0, the equation of the line L2 with respect to the line L1 symmetric is obtained

Given the line L1: x + Y-1 = 0, L2: 2x-y + 3 = 0, the equation of the line L2 with respect to the line L1 symmetric is obtained

L1:Y=1-x
L2:Y=2X+3
If two lines are symmetrical, then
Y of two analytic expressions should be equal
Then x = - 2 / 3
Straight line L passing (- 2 / 3,5 / 3)
L1 teaches X-axis at (1.0)
L2 teach X-axis at (- 3 / 2.0)
In this case, l should pass (- 1 / 4,0)
L analytic formula over (- 2 / 3,5 / 3) (- 1 / 4,0)
The solution is l: y = - 4x-1

Line y = K1X + B1 and line y = k2x + B2 (K2)

The solution of equation K1X + B1 = k2x + B2 is x = a
Inequality K1 + B1

If the line y = K1X + B1 is parallel to k2x + B2, then K1 = K2?
How to prove it?

A:
If the lines y = K1X + B1 and y = k2x + B2 are parallel, then there is no intersection, and: B1 ≠ B2
Otherwise, when x = 0, the intersection of two lines is (0, B1) = (0, B2)
Two linear equations are established
y=k1x+b1=k2x+b2
So:
(k1-k2)x=b2-b1≠0
For any real number x, the above equation has no solution
Then k1-k2 = 0
So: K1 = K2

In this paper, we use the image method to solve the inequalities (1) K1X + B1 > k2x + B2; 2) K1X + B1 < k2x + B2. First, we make the image of the functions Y1 = K1X + B1 and y2 = k2x + B2

First draw two images
k1x＋b1＞k2x＋b2
That's Y1 over Y2
So that's the range of X when Y1 is above Y2
In the same way
K1X + B1 < k2x + B2 is the range of X when Y1 is below Y2