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設P(-2y,y),則P到原點的距離的平方為OP^2=5y^2, P到直線x+2y-3=0的距離的平方為d^2=[|-2y+2y-3|^2]/(1+4)=9/5; 於是由題意可得:5y^2=9/5,得y=-3/5或y=3/5 囙此所求P點為P(6/5,-3/5)或P(-6/5,3/5).
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