在三角形ABC中BC=24,AB、AC边上的中线长之和等于39,求三角形ABC的重心的轨迹方程?
在三角形ABC中BC=24,AB、AC边上的中线长之和等于39,求三角形ABC的重心的轨迹方程?
数学人气:666 ℃时间:2019-09-16 21:47:09
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设三角形ABC重心为P,AC,AB边长中线为BE,CF\x0d由题意 BE+CF=39\x0dP是重心,有BP=(2/3)BE,CP=(2/3)CF\x0d所以BP+CP=(2/3)(BE+CF)=(2/3)*39=26>BC=24\x0dP的轨迹为以BC为焦点的长轴长为26的椭圆(除去与直线BC的交点)\x...
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