两条直线的交点坐标的综合运用1.三角形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.2.在三角形ABC中,BC边上的高所在直线的方程为x-2y+1=0,角A的平
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![两条直线的交点坐标的综合运用1.三角形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.2.在三角形ABC中,BC边上的高所在直线的方程为x-2y+1=0,角A的平](/uploads/image/z/4070220-60-0.jpg?t=%E4%B8%A4%E6%9D%A1%E7%9B%B4%E7%BA%BF%E7%9A%84%E4%BA%A4%E7%82%B9%E5%9D%90%E6%A0%87%E7%9A%84%E7%BB%BC%E5%90%88%E8%BF%90%E7%94%A81.%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.2.%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2CBC%E8%BE%B9%E4%B8%8A%E7%9A%84%E9%AB%98%E6%89%80%E5%9C%A8%E7%9B%B4%E7%BA%BF%E7%9A%84%E6%96%B9%E7%A8%8B%E4%B8%BAx-2y%2B1%3D0%2C%E8%A7%92A%E7%9A%84%E5%B9%B3)
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