সরলরেখার সমীকরণ

The angle made by the line 3xy+3=0\sqrt 3x-y+3=0 with the positive direction of X-axis is

হানি নাটস

We have in Eqny=mx+cEq^{n} y= mx+c the slope of line as m and m=tanθ,θm= \tan\theta , \theta is the angle made by (+ve) in X-axis
Hence, y=3x+3y = \sqrt{3}x+ 3
Here m=3=tanθm= \sqrt{3} =\tan\theta
θ=60o\theta=60^{o}

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