求点(2,3)关于直线l:2x-y-4=0的对称点Q的坐标解:设Q(x,y)联列: (y-3)/(x-2)=-1/2 2〔(2+x)/2〕-〔(3+y)/2〕-4=0解之得:x=22/5 y=9/5∴Q(22/5,9/5) 在百度
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![求点(2,3)关于直线l:2x-y-4=0的对称点Q的坐标解:设Q(x,y)联列: (y-3)/(x-2)=-1/2 2〔(2+x)/2〕-〔(3+y)/2〕-4=0解之得:x=22/5 y=9/5∴Q(22/5,9/5) 在百度](/uploads/image/z/14266605-21-5.jpg?t=%E6%B1%82%E7%82%B9%282%2C3%29%E5%85%B3%E4%BA%8E%E7%9B%B4%E7%BA%BFl%3A2x-y-4%3D0%E7%9A%84%E5%AF%B9%E7%A7%B0%E7%82%B9Q%E7%9A%84%E5%9D%90%E6%A0%87%E8%A7%A3%EF%BC%9A%E8%AE%BEQ%EF%BC%88x%2Cy%EF%BC%89%E8%81%94%E5%88%97%EF%BC%9A+%EF%BC%88y-3%EF%BC%89%2F%28x-2%29%3D-1%2F2+++++++++++++++++++++++++++++++++++++++++++++++++++++2%E3%80%94%282%2Bx%29%2F2%E3%80%95-%E3%80%94%EF%BC%883%2By%EF%BC%89%2F2%E3%80%95-4%3D0%E8%A7%A3%E4%B9%8B%E5%BE%97%EF%BC%9Ax%3D22%2F5++++++++++y%3D9%2F5%E2%88%B4Q%2822%2F5%2C9%2F5%29+%E5%9C%A8%E7%99%BE%E5%BA%A6)
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