陕西省2024年九年级第六次月考联考数学.考卷答案

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陕西省2024年九年级第六次月考联考数学.考卷答案试卷答案

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分析根据四边形ABCD为直角梯形需要满足的条件即可求出.

解答解:设D(m,n),A(0,-1),B(0,2),C(2,0),
则$\overline{AB}$=(0,3),$\overrightarrow{BC}$=(2,-2),$\overrightarrow{DC}$=(2-m,-n),$\overrightarrow{AD}$=(m,n+1)
当$\overrightarrow{AB}$∥$\overrightarrow{DC}$时,即3(2-m)=0,解得m=2,
且$\overrightarrow{BC}$⊥$\overline{DC}$时,即2(2-m)+2n=0,解得n=0,
满足ABCD为直角梯形.
当$\overrightarrow{AD}$∥$\overrightarrow{BC}$时,即2(n+1)=-2m,
且$\overrightarrow{AD}$⊥$\overrightarrow{AB}$时,即3(n+1)=0,解得m=0,n=-1,
满足ABCD为直角梯形.综上所述D的坐标为(2,0)或(0,-1)

点评本题主要考查两个向量共线的性质,两个向量垂直的性质,两个向量坐标形式的运算,体现了分类讨论的数学思想,属于基础题.