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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Choose the correct answer from the given four options. The equations of the lines which pass through the point (3, –2) and are inclined at $ 60^{\circ}$ to the line $3 x + y = 1$ is

$\begin {array} {1 1} (A)\;y + 2 = 0, \sqrt 3 x – y – 2 – 3 \sqrt 3 = 0 & \quad (B)\;x-2=0, \sqrt 3 x – y + 2 + 3 \sqrt 3 = 0 \\ (C)\;\sqrt 3 x – y – 2 – 3 \sqrt 3 = 0 & \quad (D)\;\text{None of these} \end {array}$

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1 Answer

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  • Equation of a line passing through a point $(x_1, y_1)$ and having slope $m$ is $ (y-y_1) = m (x-x_1)$
Step 1 :
The angle between the lines is given as $60^{\circ}$ and $ 0^{\circ}$.
Hence the slope 'm' is $ \tan 60^{\circ} = \sqrt 3 $ and $ \tan 0 = 0$
The line passing through the point (3,-2)
Hence the equation of the line is when $m =\sqrt 3 $
$ y-(-2)=\sqrt 3 (x-3)$
$ y+2=\sqrt 3x-3\sqrt 3$
$ \Rightarrow \sqrt 3x-y-2-3\sqrt 3 $
The equation of the line when $m = 0 $ is $y-(-2)=0$
$ \Rightarrow y + 2 = 0$
Hence the equation of the required lines are
$ y+2=0$ and $\sqrt 3 x-y-2-3\sqrt 3$
Hence 'A' is the correct option.
answered Jul 3, 2014 by thanvigandhi_1
 

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