The curve C is a set of points whose product of the distances from two straight lines X + y = 0 and X-Y = 0 is 1. The trajectory equation of C is obtained

The curve C is a set of points whose product of the distances from two straight lines X + y = 0 and X-Y = 0 is 1. The trajectory equation of C is obtained

The distance from point (x, y) to line x + y = 0 is | X-Y | / (√ 2)
The distance from point (x, y) to line X-Y = 0 is | x + y | / (√ 2)
So C: | X-Y | * | x + y | / 2 = 1
When x > = y, C: x ^ 2-y ^ 2 = 2
When x

The line 3x-5y + 1 = 0. Why is the linear equation of y = x symmetry 5x-3y-1 = 0
Why is the answer 5x-3y-1 = 0?

3x-5y+1=0,
y=（3x+1）/5
Let any point on the line 3x-5y + 1 = 0 be (T, (3T + 1) / 5)
Its symmetric point about the line y = x is ((3T + 1) / 5, t)
The line 3x-5y + 1 = 0 is symmetric with respect to the line y = x
y=t
x=（3t+1）/5
By eliminating T, the linear equation of the line 3x-5y + 1 = 0 with respect to the symmetry of the line y = x is 5x-3y-1 = 0
In fact, simple and intuitive approach
Choose any two points on the line 3x-5y + 1 = 0, write the symmetric points on the line y = x, and use these two points to write the linear equation, because the two points determine a straight line

What is the intersection of the line 3x + 5Y + 1 = 0 and the line 4x + 3Y + 5 = 0

3x+5y+1=0 (1)
4x+3y+5=0 (2)
(1)*4 12x+20y+4=0
(2) * 3 12x + 9y + 15 = 0 minus
11y-11=0
y=1
Substitute (1) to get x = - 2
The intersection is (- 2,1)