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Question

A straight line L through the point $$(3,-2)$$ is inclined at an angle $$60$$ to the line $$\sqrt { 3 } x+y=1$$. If L also intersects the x-axis, the equation of L is-


A
y+3x+233=0
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B
y3x+2+33=0
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C
3yx+3+23=0
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D
3y+x3+23=0
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Solution

The correct option is D $$y-\sqrt { 3 }x+2+3\sqrt { 3 }=0$$
$$\sqrt{3}x+y=1\implies m=\tan \theta =-\sqrt{3}\implies $$ this line makes an angle of $$120^{\circ}$$ with x-axis.

Since $$ L$$ makes an angle $$60^{\circ}$$ with this line.

$$\implies L$$ must be either x- axis $$($$ which is not correct because given $$L$$ intersect the x-axis$$)$$ or a line making an angle $$60^{\circ}$$ with the x-axis.

$$\implies m=\tan 60^{\circ}=\sqrt{3}$$ 

$$L:y=m{x}+c$$

$$L:y=\sqrt{3}x+c$$
 
$$L$$ passes through $$(3,-2)\implies -2=3\sqrt{3}+c\implies c=-2-3\sqrt{3}$$

$$L:y-\sqrt{3}x+2+3\sqrt{3}=0$$ 

Mathematics

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