设 $\triangle ABC$ 三个顶点的坐标分别为 $A\left( {2,1} \right)$，$B\left( { - 1, 2} \right)$，$C\left( {3, - 1} \right)$，$D,E$ 分别为 $AB,BC$ 上的点，$M$ 是 $DE$ 上一点，且 $\dfrac{{BE}}{{BC}} = \dfrac{{AD}}{{AB}} = \dfrac{{DM}}{{DE}}$．
【难度】
【出处】
2009年华南理工大学自主招生保送生选拔考试
【标注】
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求点 $M$ 的横坐标的取值范围；标注答案$\left[ {\dfrac{5}{7},3} \right)$解析设 $\dfrac{{BE}}{{BC}} = \dfrac{{AD}}{{AB}} = \dfrac{{DM}}{{DE}} = \lambda $，则 $0 < \lambda < 1$ 且\[\begin{split}\overrightarrow {OD} &= \lambda \overrightarrow {OB} + \left( {1 - \lambda } \right)\overrightarrow {OA} ,\\ \overrightarrow {OE} &= \lambda \overrightarrow {OC} + \left( {1 - \lambda } \right)\overrightarrow {OB} ,\\\overrightarrow {OM} &= \lambda \overrightarrow {OE} + \left( {1 - \lambda } \right)\overrightarrow {OD},\end{split}\]于是$$\overrightarrow {OM} = {\lambda ^2}\overrightarrow {OC} + 2\lambda \left( {1 - \lambda } \right)\overrightarrow {OB} + {\left( {1 - \lambda } \right)^2}\overrightarrow {OA} = \left( {7{\lambda ^2} - 6\lambda + 2, - 4{\lambda ^2} + 2\lambda + 1} \right).$$所以其横坐标$$7{\lambda ^2} - 6\lambda + 2 = 7{\left( {\lambda - \dfrac{3}{7}} \right)^2} + 2 - \dfrac{9}{7} = 7{\left( {\lambda - \dfrac{3}{7}} \right)^2} + \dfrac{5}{7},$$取值范围是 $ \left[ {\dfrac{5}{7} ,3} \right)$．
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求点 $M$ 的轨迹方程．标注答案$16x^2+56xy+49y^2-100x-150y+125=0$解析设点 $M\left( {x , y} \right)$ 则$$x = 7{\lambda ^2} - 6\lambda + 2, y = - 4{\lambda ^2} + 2\lambda + 1,$$由$$4x + 7y = 15 - 10\lambda ,$$解出$$\lambda = \dfrac{{15 - 4x - 7y}}{{10}},$$代入$$x = 7{\lambda ^2} - 6\lambda + 2,$$可得点 $M$ 的轨迹方程为$${\left( {15 - 4x - 7y} \right)^2} = - 20x - 60y + 100,$$也即$$16x^2+56xy+49y^2-100x-150y+125=0.$$
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