设B点为(x,y)
(y-y1)/(x-x1) = -b/a; a*(x+x1)/2+b*(y+y1)+c = 0
化简得 bx+ay=bx1+ay1; (1)
ax+by=-ax1-by1-2c (2)
a*(1)-b*(2) 得 y=[(a^2+b^2)*y1+2ab*x1+2bc]/(a^2-b^2);
b*(1)-a*(2) 得 x=-[(a^2+b^2)*x1+2ab*y1+2ac]/(a^2-b^2)
设B点为(x,y)
(y-y1)/(x-x1) = -b/a; a*(x+x1)/2+b*(y+y1)+c = 0
化简得 bx+ay=bx1+ay1; (1)
ax+by=-ax1-by1-2c (2)
a*(1)-b*(2) 得 y=[(a^2+b^2)*y1+2ab*x1+2bc]/(a^2-b^2);
b*(1)-a*(2) 得 x=-[(a^2+b^2)*x1+2ab*y1+2ac]/(a^2-b^2)