点c坐标是(2,2)过点C的直线CA与x轴交于A,过点C且与直线CA垂直的直线CB与y轴交于点B,设M是AB的中点,求M...
问题描述：
点c坐标是(2,2)过点C的直线CA与x轴交于A,过点C且与直线CA垂直的直线CB与y轴交于点B,设M是AB的中点,求M...
点c坐标是(2,2)过点C的直线CA与x轴交于A,过点C且与直线CA垂直的直线CB与y轴交于点B,设M是AB的中点,求M的轨迹方程
答
设直线是y-2=k(x-2)y=0,x=-2/k+2=(2k-2)/k所以A[(2k-2)/k,0]BC斜率-1/ky-2=-1/k(x-2)x=0y=2+2/k=(2k+2)/kB[0,(2k+2)/k]中点x=[(2k-2)/k+0]/2=(k-1)/ky=[0+(2k+2)/k]/2=(k+1)/k相加x+y=2