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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Find the equation of lines passing through (1, 2) and making angle $30^{\circ}$ with $y$ - axis.

$\begin {array} {1 1} (A)\;y+\sqrt 3 x - 2 - \sqrt 3=0 & \quad (B)\;y-\sqrt 3 x + 2 + \sqrt 3=0 \\ (C)\;y-\sqrt 3 x - 2 + \sqrt 3=0 & \quad (D)\;y-\sqrt 3 x - 2 - \sqrt 3=0 \end {array}$

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1 Answer

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Step 1 :
It is given that the line makes an angle of $30^{\circ}$ with $y$ - axis.
$ \therefore $ The angle made with $x$ axis in the positive direction is $60^{\circ}$
$ \therefore \tan 60^{\circ} = \sqrt 3$
Hence slope $m = \sqrt 3$
It is given that it passes through the point (1, 2)
Step 2 :
Hence the equation of the line is
$ (y-2)=\sqrt 3 (x-1)$
$ \Rightarrow y-2 = \sqrt 3x - \sqrt 3$
(i.e) $y-\sqrt 3 x - 2+ \sqrt 3 = 0$
Hence the equation of the required line is
$ y-\sqrt 3 x - 2+\sqrt 3 = 0$
answered Jul 1, 2014 by thanvigandhi_1
 

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