Find the symmetric point coordinates of point P (x0, Y0) with respect to AX + by + C = 0

Find the symmetric point coordinates of point P (x0, Y0) with respect to AX + by + C = 0

Let P (x1, Y1) when a = 0, we can solve X1 = x0, Y1 = (- 2C / b) - Y0; when B = 0, we can solve X1 = (- 2C / a) - x0, Y1 = Y0; when AB is not equal to 0, because it is a symmetric point, the midpoint of the line segment is on the straight line: a (x1 + x0) / 2 + B (Y1 + Y0) / 2 + C = 0, vertical: (y1-y0) / (x1-x0) * - A / b = - 1, we can solve X1 = - (2ca + aby0-x0

Let the equation of l be ax + by + C = 0 (a ^ 2 + B ^ 2 ≠ 0) and the point P (x0, Y0) be known, then the linear equation of L with respect to p is obtained
Let p '(x', y ') be any point on the symmetric line L', and his symmetric point (2x0-x ', 2y0-y') on P (x0, Y0) be substituted on the line L
A (2x0-x ') + B (2y0-y') + C = 0 is the symmetric linear equation
Why do I substitute the equation of a straight line L to get the equation of L '?

L^:Ax+By+c=0
Let l be the symmetric point (x, y) of P (x0, Y0)
x(L)+x(L^)=2x0,x(L^)=2x0-x(L)=2x0-x
y(L)+y(L^)=2y0,y(L^)=2y0-(y(L)=2y0-y
L^:Ax^+By^+c=0
A*(2x0-x)+B*(2y0-y)+C=0

As shown in the figure below, the line AB passes through the point C on the circle O, and OA = ob, CA = CB. Prove that the line AB is the tangent of the circle o

Certification:
Connect to OC
∵OA=OB,CA=CB,OC=OC
∴⊿AOC≌⊿BOC（SSS）
∴∠ACO=∠BCO
∵∠ACO+∠BCO=180º
∴∠ACO=∠BCO=90º
That is OC ⊥ AB, according to the straight line perpendicular to the outer end of the radius is the tangent of the circle
The line AB is tangent to the circle o