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設該直線的斜率為k,有直線過點(-2,2),則直線的方程為y=kx+2(k+1). 當x=0時,y=2(k+1).當y=0時,x=-2(k+1)/k S=1/2*x*y=-1/2*2(k+1)*2(k+1)/k=1 解得k=-1/2或k=-2 所以該直線的方程為y=-1/2+1,或y=-2x-2
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