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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
人气:234 ℃ 时间:2020-05-25 07:11:24
解答
斜率分别是k和-1/k则L1y-1=k(x-3)y=0,x=(3k-1)/kL2y-1=-(1/k)(x-3)x=0y=(k+3)/k所以A((3k-1)/k,0),B(0,(k+3)/k)中点则x=[(3k-1)/k+0]/2y=[0+(k+3)/k]/22x=(3k-1)/k2y=(k+3)/k所以2x-6y=(3k-1-3k-9)/k=-10/k因为y-1=k...
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