1樓：匿名使用者
function which(byval p as pointf) as string
dim tp as string = math.sign(p.x) & "," & math.sign(p.y)
select case tp
case "0,0"
return "原點"
case "0,1", "0,-1"
return "y軸"
case "1,0", "-1,0"
return "x軸"
case "1,1"
return "第一象限"
case "-1,1"
return "第二象限"
case "-1,-1"
return "第三象限"
case "1,-1"
return "第四象限"
end select
end function
2樓：匿名使用者
dim inx as long ,iny as longinx = val(inputbox("請輸入x軸數值"))iny = val(inputbox("請輸入y軸數值"))if inx>0 then
if iny>0 then 'x,y >0 第一象限msgbox "座標(" & inx & "," & iny & ") 在第一象限"
elseif iny<0 then ' x> 0 >y第四象限msgbox "座標(" & inx & "," & iny & ") 在第四象限"
end if
elseif inx<0 then
if iny>0 then 'y>0>x 第二象限msgbox "座標(" & inx & "," & iny & ") 在第二象限"
elseif iny<0 then ' x0 thenmsgbox "座標在x軸上"
end if
end if
if iny=0 then
if inx<>0 then
msgbox "座標在y軸上"
end if
end if
'懶得開vb了
（用c++編寫程式）輸入平面直角座標系中一點的座標（x,y），判斷改點是在那個象限中或那一條座標軸上
3樓：匿名使用者
include "math.h"
class point
};class line
float getdistance(point p)};void main()
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