Given that the line L1 = 2x + 3, the line L2 and L1 are symmetric with respect to the line y = - x, then the slope of the line L2 is?

Given that the line L1 = 2x + 3, the line L2 and L1 are symmetric with respect to the line y = - x, then the slope of the line L2 is?

-Half of it

Find the equation of the line L 1: y = 2x + 3 with respect to the symmetric line L 2 of the line L 1: y = x + 1

Take any point a (x, y) on the line L2, then the symmetric point B (x ', y') of point a about the line L1: y = x + 1 is on the line L1: y = 2x + 3, from y '- YX' - x = - 1, and y '+ y2 = x + X' 2 + 1, we can get: X '= Y-1, y' = x + 1, ∧ B (Y-1, x + 1), ∧ x + 1 = 2 (Y-1) + 3, that is, the equation of the line L2 is & nbsp; x-2y = 0

It is known that the line L: x-y-1 = 0, L1: 2x-y-2 = 0. If the line L2 and L1 are symmetric with respect to L, then the equation of L2 is?
But there is a solution that you can't understand
If x-y-1 = 0, then x = y + 1, y = X-1, then the L1 equation 2x-y-2 = 0, then the L2 equation 2 (y + 1) - (x-1) - 2 = 0, that is, x-2y-1 = 0

Let a (x, y) be the point on the L2 equation
Then the symmetric point of a with respect to L1 is B (y + 1, x-1), and f (x, y) = 0, which is the equation of L1
Think of it as an equation
This equation represents the linear system formed by the rotation of a straight line with point a as the fixed point
Because two points determine a straight line, so as long as the B point into the equation

Given the line L1: x + Y-1 = 0, L2: 2x-y + 3 = 0, find the equation of the line L2 with respect to the line L1 symmetric
From the solution of L1 and L2, the intersection point is x = - 2 / 3, y = 5 / 3
Go to a point Q (- 1,1) on L2
Let the symmetric point of Q with respect to line L1 be Q1 (x1, Y1)
One is (x1-1) / 2 + (Y1 + 1) / 2-1 = 0
One is y1-1 / x1-1 times (- 1) = - 1
What I want to ask is that (x1-1) / 2 (Y1 + 1) / 2 should be the midpoint on L. why is it brought into L1

This axis of symmetry is special, that is, if the angle between the axis of symmetry and the X axis is 45 degrees, the following method can be used:
Let's write one equation of the axis of symmetry into two,
The symmetry equation can be written as:
{y=1-x
{x=1-y
Replace X in L2 with: (1-y)
Y is replaced by: (1-x)
The results are as follows
2(1-y)-(1-x)+3=0
Namely:
x-2y+4=0
The above evidence comes from the translation of function and inverse function plus coordinate system