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AB=5,三角形ABC面積=ABd/2=5,d=2 點C在與直線AB平行且距離為2的直線上,這樣的直線有兩條 AB的方程為:4x-3y+2=0 設C(x,y),則由點到直線距離公式得: 4x-3y+2/5=2 即點C軌跡方程為:4x-3y-8=0,或4x-3y+14=0
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