过定点A(2,4)任作互相垂直的两条线L1与L2,设L1与x轴交于点M,L2与y轴交于点N,求线段MN的中点P的轨迹方程.

设M(m,0),N(0,n)向量MA= (m-2,-4)向量MB = (-2,n-4)向量MB垂直向量MA:-2(m-2) -4(n-4) = 0m = 10 - 2nM(10-2n,0),N(0,n)线段MN的中点P(x,y):x = (10 -2n)/2 = 5 - ny = n/2消去n,x+2y -5 = 0