设所求直线方程L:4x+3y+b=0
x=0,y=-b/3
y=0,x=-b/4
√[(-b/3)^2+(-b/4)^2]+|-b/3|+|-b/4|=10
|5b/12|+|7b/12|=10
|b|=10
b=±10
答:所求直线方程有两条,即:4x+3y±10=0