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由題意可得,可設直線l的方程為y=16x+b,顯然此直線和兩坐標軸的交點分別為(0,b)、(-6b,0).再由直線和兩坐標軸圍成面積為3的三角形,可得12|b|•|-6b|=3,解得 ;b=±1,故直線的方程為y =16x±1,即x-…
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