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當直線過原點時,直線方程為y=43x,即4x-3y=0;當直線不過原點時,設直線方程為x+y=a.則3+4=a,得a=7.∴直線方程為x+y-7=0.∴過點M(3,4)且在坐標軸上截距相等的直線方程為4x-3y=0或x+y-7=0.故答案為:4x-3y=0或x+y-7=0.
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