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4x+3y-10=0 設直線為y=kx+b,由已知條件可求出,斜率k=負的三分之四,再設x=0和y=0是求出與坐標軸的交點,確定截距,最後是b+四分之三b+根號下b^2+四分之三b^2=10,可以求出b=三分之十
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