每日一题:2020-04-15

每日一题: 2020-04-15

题目:
已知A(3,2),B(2,3),C(3,1)A(-3,2), B(-2,-3), C(3,1), 试求ABC\triangle ABC 的重心坐标及外心坐标.

参考思路

如图所示, 设D,ED,E 分别为BC,ACBC,AC 的中点, 所以D(12,1),E(0,32)D(\frac{1}{2},-1), E(0,\frac{3}{2}).
易求得AD:y=67x47;BE:y=94x+32AD: y=-\frac{6}{7}x-\frac{4}{7}; BE: y=\frac{9}{4}x+\frac{3}{2}. 联立得
\[
\left\{\begin{array}{lr} y=-\frac{6}{7}x-\frac{4}{7} \\ y=\frac{9}{4}x+\frac{3}{2} \end{array}\right.\Rightarrow G(-\frac{2}{3},0)
\]

再易求得BC:y=45x75;AC:y=16x+32BC: y=\frac{4}{5}x-\frac{7}{5}; AC: y=-\frac{1}{6}x+\frac{3}{2}.
所以设BCBC 的垂直平分线OD:y=54x+b1OD: y=-\frac{5}{4}x+b_1; ACAC 的垂直平分线OE:y=6x+b2OE:y=6x+b_2.
代入D(12,1),E(0,32)D(\frac{1}{2},-1), E(0,\frac{3}{2}) 解得b1=38,b2=32b_1=-\frac{3}{8}, b_2=\frac{3}{2}

\[
\left\{\begin{array}{lr} y=-\frac{5}{4}x-\frac{3}{8} \\ y=-\frac{1}{6}x+\frac{3}{2} \end{array}\right.\Rightarrow O(-\frac{15}{58},-\frac{3}{58})
\]

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