设点p(m,n)关于直线kx-y+b=0的对称点p'(m',n') (n'-n)/(m'-m)=-1/k→m'=m+k(n-n').|km-n+b|/√(k+1)=d=|km'-n'+b|/√(k+1)=|k[m+k(n-n')]-n'+b]/√(k+1)→n'=[2km+(k-1)n+2b)/(k+1)→ m'=(2kn-km+2kb+m)/(k+1)
设点p(m,n)关于直线kx-y+b=0的对称点p'(m',n') (n'-n)/(m'-m)=-1/k→m'=m+k(n-n').|km-n+b|/√(k+1)=d=|km'-n'+b|/√(k+1)=|k[m+k(n-n')]-n'+b]/√(k+1)→n'=[2km+(k-1)n+2b)/(k+1)→ m'=(2kn-km+2kb+m)/(k+1)