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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Find the equation of a straight line on which length of perpendicular from the origin is four units and the line makes an angle of $120^{\circ}$ with the positive direction of $x$ - axis.\[\] [Hint: Use normal form, here $ \omega =30^{\circ}$.]

$\begin {array} {1 1} (A)\;\sqrt 3x+y=8 & \quad (B)\;x+y\sqrt 3 =8 \\ (C)\;\sqrt 3 x-y=8 & \quad (D)\;x-y\sqrt 3=8 \end {array}$

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  • Equation of the line in the normal form is $ x \cos \omega + y \sin \omega = p$
Step 1 :
Given $p=4$ units and $ \alpha = 120^{\circ}$
hence $\omega = 30^{\circ}$
$ \therefore $ Equation of the line in the normal form is
$x \cos 30^{\circ}+y \sin 30^{\circ}=4$
$ \cos 30^{\circ}= \large\frac{\sqrt 3}{2}$ and $ \sin 30^{\circ}= \large\frac{1}{2}$
$ \therefore x\large\frac{\sqrt 3}{2}$$+\large\frac{y}{2}$$=4$
(i.e) $ \sqrt 3x+y=8$ is the required equation.
answered Jul 1, 2014 by thanvigandhi_1

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