Given the line I: y = 3x + 3, the equation of the line x-y-2 = 0 with respect to the l-symmetric line is obtained From x-y-2 = 0 and 3x-y + 3 = 0, x = - 5 / 2, y = - 9 / 2 Then three straight lines intersect at the point: (- 5 / 2, - 9 / 2) Let the linear equation be y + 9 / 2 = K (x + 5 / 2) The angle from the straight line x-y-2 = 0 to the straight line 3x-y + 3 = 0 is equal to the angle from the straight line 3x-y + 3 = 0 to the straight line (3-1)/(1+3),=(K-3)/(1+3k) So, k = - 7 So the linear equation is: 7x + y + 22 = 0 Why is the angle from the straight line x-y-2 = 0 to the straight line 3x-y + 3 = 0 equal to the angle from the straight line 3x-y + 3 = 0 to the straight line (3-1)/(1+3),=(K-3)/(1+3k)

Given the line I: y = 3x + 3, the equation of the line x-y-2 = 0 with respect to the l-symmetric line is obtained From x-y-2 = 0 and 3x-y + 3 = 0, x = - 5 / 2, y = - 9 / 2 Then three straight lines intersect at the point: (- 5 / 2, - 9 / 2) Let the linear equation be y + 9 / 2 = K (x + 5 / 2) The angle from the straight line x-y-2 = 0 to the straight line 3x-y + 3 = 0 is equal to the angle from the straight line 3x-y + 3 = 0 to the straight line (3-1)/(1+3),=(K-3)/(1+3k) So, k = - 7 So the linear equation is: 7x + y + 22 = 0 Why is the angle from the straight line x-y-2 = 0 to the straight line 3x-y + 3 = 0 equal to the angle from the straight line 3x-y + 3 = 0 to the straight line (3-1)/(1+3),=(K-3)/(1+3k)

Because 3x-y + 3 = 0 is the axis of symmetry of two symmetrical lines, it can be proved from the knowledge of plane geometry that the axis of symmetry is the angular bisector of two axisymmetric lines
According to the relationship between the slope and the tangent of the inclination angle, and the tangent formula of the difference between the two angles, let the inclination angle of the original line be α, the inclination angle of the symmetrical axis line be β, and the inclination angle of the axisymmetric line be γ
Then K primitive = 1; K axis = 3; K solution = k = > Tan α = 1; Tan β = 3; Tan γ = K
And β - α = γ - β = > Tan (β - α) = Tan (γ - β) = > (Tan β - Tan α) / (1 + Tan α, Tan β) = (Tan γ - Tan β) / (1 + Tan γ, Tan β)
Substituting the corresponding values, we can get the formula given in the question