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AB=√[(-3)²;+(-4)²;]=5 直線AB所在的方程為4x-3y=0 點C(4,-2)到直線AB的距離為 |16+6|/√(3²;+4²;)=22/5 所以 三角形ABC的面積為 S=(1/2)*5*(22/5)=11
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