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既然是和坐標軸正半軸有交點,則此直線的斜率k0,b>0,且ab=10 又直線過(-5,4),則(-5)/a+4/b=1 ===>>>-5b+4a=ab=10 ==>> 4a-5b=10 ab=10 解得a=5,b=2 所求直線方程是x/5+y/2=1即:2x+5y-10=0
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