无论k为何值,直线(2k+1)x-(k-2)y-(k+8)=0恒过一个定点,则这个定点是?(2k+1)x-(k-2)y-(k+8)=02xk+x-yk+2y-k-8=0(2x-y-1)k+(x+2y-8)=0(2x-y-1)=0(x+2y-8)=0x=2 y=3点(2,3) 其中,为什么要让 (2x-y-1)=0 那么,为
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![无论k为何值,直线(2k+1)x-(k-2)y-(k+8)=0恒过一个定点,则这个定点是?(2k+1)x-(k-2)y-(k+8)=02xk+x-yk+2y-k-8=0(2x-y-1)k+(x+2y-8)=0(2x-y-1)=0(x+2y-8)=0x=2 y=3点(2,3) 其中,为什么要让 (2x-y-1)=0 那么,为](/uploads/image/z/12517921-1-1.jpg?t=%E6%97%A0%E8%AE%BAk%E4%B8%BA%E4%BD%95%E5%80%BC%2C%E7%9B%B4%E7%BA%BF%EF%BC%882k%2B1%EF%BC%89x-%EF%BC%88k-2%EF%BC%89y-%EF%BC%88k%2B8%EF%BC%89%3D0%E6%81%92%E8%BF%87%E4%B8%80%E4%B8%AA%E5%AE%9A%E7%82%B9%2C%E5%88%99%E8%BF%99%E4%B8%AA%E5%AE%9A%E7%82%B9%E6%98%AF%3F%EF%BC%882k%2B1%EF%BC%89x-%EF%BC%88k-2%EF%BC%89y-%EF%BC%88k%2B8%EF%BC%89%3D02xk%2Bx-yk%2B2y-k-8%3D0%282x-y-1%29k%2B%28x%2B2y-8%29%3D0%282x-y-1%29%3D0%28x%2B2y-8%29%3D0x%3D2+y%3D3%E7%82%B9%282%2C3%29+%E5%85%B6%E4%B8%AD%2C%E4%B8%BA%E4%BB%80%E4%B9%88%E8%A6%81%E8%AE%A9+%282x-y-1%29%3D0+%E9%82%A3%E4%B9%88%EF%BC%8C%E4%B8%BA)
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