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  • A straight line L through the point (3, -2) is inclined at an angle of 60o to the line <img alt="\sqrt{3}\; x+y=1" src="https://entrancecorner.oncodecogs.com/gif.la

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

 

Option: 1

y+\sqrt{3}\; x+2-3\sqrt{3}=0


Option: 2

y-\sqrt{3}\; x+2+3\sqrt{3}=0


Option: 3

\sqrt{3}\;y- x+3+2\sqrt{3}=0


Option: 4

\sqrt{3}\;y+ x-3+2\sqrt{3}=0


Answers (1)

best_answer

Let the slope of line L be m

Slope of given line (m1)=-\sqrt{3}

Angle between them =60^{\circ}

\tan 60^{\circ}=\left | \frac{m+\sqrt{3}}{1-\sqrt{3}m} \right |

\frac{m+\sqrt{3}}{1-\sqrt{3}m} =\pm \sqrt{3}

On solving m=0 \: or \: m=\sqrt{3}

And these lines pass through (3, -2)

Using point-slope form, we get  y+2 = 0 or \sqrt{3}x-y-3\sqrt{3}-2=0

But y + 2 = 0 does not intersect x-axis as it is parallel to it, so th eother line is the answer

Posted by

Devendra Khairwa

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