Question

L1与L2过点M（3,1）并且互相垂直分别交X 轴A Y轴B 线段AB中点Q 求Q的移动轨迹

Answer 1

斜率分别是k和-1/k

则L1

y-1=k(x-3)

y=0,x=(3k-1)/k

L2

y-1=-(1/k)(x-3)

x=0

y=(k+3)/k

所以A((3k-1)/k,0),B(0,(k+3)/k)

中点则x=[(3k-1)/k+0]/2

y=[0+(k+3)/k]/2

2x=(3k-1)/k

2y=(k+3)/k

所以2x-6y=(3k-1-3k-9)/k=-10/k

因为y-1=k(x-3)

k=(y-1)/(x-3)

所以(2x-6y)(y-1)=-10(x-3)

2xy-2x-6y²+6y=-10x+30

3y²-xy-4x-3y+15=0