P（x,y）和P'(x',y')关于直线y=kx对称。求相应的坐标变换公式和对应的二阶矩阵。
\u5199\u51fa\u5e73\u9762\u4e0a\u7684\u70b9\uff08x,y\uff09\u5bf9\u4efb\u610f\u76f4\u7ebfy=kx+b\u4f5c\u5bf9\u79f0\u53d8\u6362\u7684\u53d8\u6362\u77e9\u9635\uff1f\u6025\u9700\u5176\u7b54\u6848\uff01\u6025\u6025\u6025\u6025\u6025\u6025\uff01\u76f4\u7ebfy=kx+b,k\u662f\u975e0\u7684\u6709\u9650\u503c\uff0c\u4efb\u610f\u70b9\uff08m\uff0cn\uff09 \u3002
\u5bf9\u79f0\u70b9\u5728\u8fc7\u5df2\u77e5\u70b9\uff0c\u5e76\u5782\u76f4\u4e8e\u5df2\u77e5\u76f4\u7ebf\u4e0a\u3002
\u901a\u8fc7\u89e3\u65b9\u7a0b\u7ec4
y=kx+b
y-n=(-1/k)*(x-m)
\u5f97\u5230\u5782\u70b9\uff0c\u4e5f\u5c31\u662f\u4e24\u70b9\u7684\u4e2d\u70b9\u5750\u6807\u4e3a\uff1a
x= (m+kn-kb) /(k*k+1)
y=(km+k*kn+b)/(k*k+1)
\u4ece\u800c\u5f97\u5230\u5bf9\u79f0\u70b9\u5750\u6807\u4e3a\uff1a
x= 2(m+kn-kb) /(k*k+1)-m
y=2(km+k*kn+b)/(k*k+1)-n
\u4ece\u9898\u76ee\u6765\u770b\uff0c(a,b)\u70b9\u521a\u597d\u5728\u76f4\u7ebfy=kx+b\u4e0a\uff0c\u5373(0,b)\u70b9\uff0c\u90a3\u4e48\u5173\u4e8ey=kx+b\u7684\u5bf9\u79f0\u70b9\u5219\u662f\u5176\u81ea\u8eab\u3002
解：(x+x')/2*k=(y+y')/2,且(y-y')/（x-x')=-1/k 所以：有y'=（(k2+1）y-2kx)/(k2-1) x'=（2ky-(k2+1）x)/(k2-1)变换矩阵为：M= 2k -( k2+1) k2+1 -2k教你方法，在原直线上取两点，将点关于直线对称过去，再算所求直线
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