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Find the equation of the line which satisfy the given conditions: 

Q : 4         Passing through  (2,2\sqrt{3})  and inclined with the x-axis at an angle of 75^{\circ}.

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We know that the equation of the line passing through the point (x_1,y_1) and with slope m is given by
(y-y_1)=m(x-x_1)
we know that
m = \tan \theta
where \theta is angle made by line with positive x-axis measure in the anti-clockwise direction
m = \tan75\degree \ \ \ \ \ \ \ \ \ \ \ \ \ (\because \theta=75\degree \ given)
m = \frac{\sqrt3+1}{\sqrt3-1}
Now, the equation of the line passing through the point (2,2\sqrt3) and with slope m = \frac{\sqrt3+1}{\sqrt3-1}  is 
(y-2\sqrt3)=\frac{\sqrt3+1}{\sqrt3-1}(x-2)\\ \\ (\sqrt3-1)(y-2\sqrt3)=(\sqrt3+1)(x-2)\\ (\sqrt3-1)y-6+2\sqrt3= (\sqrt3+1)x-2\sqrt3-2\\ (\sqrt3+1)x-(\sqrt3-1)y = 4(\sqrt3-1)
Therefore, the equation of the line  is   (\sqrt3+1)x-(\sqrt3-1)y = 4(\sqrt3-1)

Posted by

Gautam harsolia

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