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設直線為y-2=k(x+2),交x軸於點(−2k−2,0),交y軸於點(0,2k+2),S=12×|2k+2|×|2k+2|=1,|4+2k+2k|=1得2k2+3k+2=0,或2k2+5k+2=0解得k=−12,或k=-2,∴x+2y-2=0,或2x+y+2=0為所求.
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