三角形ABC的顶点A的坐标为(1,4),角B,角C平分线的方程分别为x-2y=0和x+y-1=0,求BC边所在的直线方程用对称做
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![三角形ABC的顶点A的坐标为(1,4),角B,角C平分线的方程分别为x-2y=0和x+y-1=0,求BC边所在的直线方程用对称做](/uploads/image/z/968487-15-7.jpg?t=%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E9%A1%B6%E7%82%B9A%E7%9A%84%E5%9D%90%E6%A0%87%E4%B8%BA%EF%BC%881%2C4%EF%BC%89%2C%E8%A7%92B%2C%E8%A7%92C%E5%B9%B3%E5%88%86%E7%BA%BF%E7%9A%84%E6%96%B9%E7%A8%8B%E5%88%86%E5%88%AB%E4%B8%BAx-2y%3D0%E5%92%8Cx%2By-1%3D0%2C%E6%B1%82BC%E8%BE%B9%E6%89%80%E5%9C%A8%E7%9A%84%E7%9B%B4%E7%BA%BF%E6%96%B9%E7%A8%8B%E7%94%A8%E5%AF%B9%E7%A7%B0%E5%81%9A)
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