发布网友 发布时间:2024-10-03 02:06
共1个回答
热心网友 时间:2024-10-03 02:48
解:(1)解 得(x﹣https://iknow-pic.cdn.bcebos.com/b543a98226cffc1e178a82b9014a90f603eac4?x-bce-process=image/quality,q_85 )(x﹣1)=0,
解得x 1 =https://iknow-pic.cdn.bcebos.com/b543a98226cffc1e178a82b9014a90f603eac4?x-bce-process=image/quality,q_85 ,x 2 =1。
∵OA<OB,∴OA=1,OB=https://iknow-pic.cdn.bcebos.com/b543a98226cffc1e178a82b9014a90f603eac4?x-bce-process=image/quality,q_85 。∴A(1,0),B(0,https://iknow-pic.cdn.bcebos.com/b543a98226cffc1e178a82b9014a90f603eac4?x-bce-process=image/quality,q_85 )。∴AB=2。
又∵AB:AC=1:2,∴AC=4。∴C(﹣3,0)。;
(2)由题意得:CM=t,CB=2https://iknow-pic.cdn.bcebos.com/b543a98226cffc1e178a82b9014a90f603eac4?x-bce-process=image/quality,q_85 .
①当点M在CB边上时,S=2https://iknow-pic.cdn.bcebos.com/b543a98226cffc1e178a82b9014a90f603eac4?x-bce-process=image/quality,q_85 ﹣t(0≤t<https://iknow-pic.cdn.bcebos.com/b543a98226cffc1e178a82b9014a90f603eac4?x-bce-process=image/quality,q_85 );
②当点M在CB边的延长线上时,S=t﹣https://iknow-pic.cdn.bcebos.com/b543a98226cffc1e178a82b9014a90f603eac4?x-bce-process=image/quality,q_85 (t>https://iknow-pic.cdn.bcebos.com/b543a98226cffc1e178a82b9014a90f603eac4?x-bce-process=image/quality,q_85 )。
(3)存在,Q 1 (﹣1,0),Q 2 (1,﹣2),Q 3 (1,2),Q 1 (1,https://iknow-pic.cdn.bcebos.com/b543a98226cffc1e178a82b9014a90f603eac4?x-bce-process=image/quality,q_85 )。
试题分析:(1)通过解一元二次方程 ,求得方程的两个根,从而得到A、B两点的坐标,再根据勾股定理可求AB的长,根据AB:AC=1:2,可求AC的长,从而得到C点的坐标。
(2)分①当点M在CB边上时;②当点M在CB边的延长线上时;两种情况讨论可求S关于t的函数关系式。
(3)分AB是边和对角线两种情况讨论可求Q点的坐标: