已知△ABC三个顶点坐标分别为A（1，2），B（3，1），C（-1，-1）．
（1）求BC边上的高线所在的直线方程；
（2）求BC边上的中线所在的直线方程．
（1）由题意可得直线BC的斜率kBC=
=-1-1 -1-3
，1 2
∴BC边上的高线所在的直线的斜率为-2，
∴所求直线的方程为：y-2=-2（x-1），
化为一般式可得：2x+y-4=0
（2）∵B（3，1），C（-1，-1），
∴BC的中点D的坐标为（1，0），
∴BC边上的中线所在的直线方程为：x=1
| 从中选出适当的词组填空。 | |
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