Find the linear equation with the slope of - 1 / 2 and passing through the intersection of two lines 2x + 3y-3 = 0 and X-Y = 0

Find the linear equation with the slope of - 1 / 2 and passing through the intersection of two lines 2x + 3y-3 = 0 and X-Y = 0

y=x
2x+3x-3=0
x=3/5
The intersection of two straight lines is (3 / 5,3 / 5)
So the linear equation is Y-3 / 5 = - (1 / 2) (x-3 / 5)
Multiply by 10
10y-6=-5x+3
That is, 5x + 10y-9 = 0

Find the N equation of the line M: 2x + y-4 = 0 symmetric to the line L: 3x + 4y-1 = 0

Take any point on the line m, such as (0,4), and let its symmetric point on the line l be (x, y), then the midpoint of these two points [x, (y + 4) / 2] is on the line L, and substitute it into the equation of L to get 3x + 2Y + 7 = 0

Solving the equation of the line 2x + Y-3 = 0 with respect to the symmetric line 3x + 4Y + 4 = 0

The equation of 2x + Y-3 = 0 for the symmetric line 3x + 4Y + 4 = 0 is as follows:
（3x+4y+4）- （2x+y-3）=0
That is: x + 3Y + 7 = 0