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A straight line L through the point (3, -2) is inclined at an angle of 600 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

As learnt in concept

Angle between two lines (θ) -

\tan \Theta = \frac{m_{2}-m_{1}}{1+m_{1}m_{2}}

- wherein

Here m_{1},m_{2} are the slope of two lines

 

 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} =^+_-\sqrt{3}

on solving m=0 \: or \: m=\sqrt{3}

And these lines pass through (3, -2)

we get  y+2 = 0

or \sqrt{3}x-y-3\sqrt{3}-2=0


Option 1)

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

Incorrect option    

Option 2)

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

Correct option

Option 3)

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

Incorrect option    

Option 4)

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

Incorrect option    

Posted by

divya.saini

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