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Line L_{1} is perpendicular to \sqrt{3}x+y= 0. If L_{2} makes an angle of 45^{\circ} with L_{1}, then the slope of L_{2} can be

Option: 1

2\; or\; \frac{1}{2}


Option: 2

3\; or\; \frac{1}{3}


Option: 3

4\; or\; \frac{1}{4}


Option: 4

None of these


Answers (1)

best_answer

Slope\: of\: \sqrt{3}x+y= 0 \Rightarrow y= -\sqrt{3}x
slope= -\sqrt{3}= m\left ( say \right )

As product of slopes of perpendicular lines is -1, so slope of L_{1}\left ( m_{1} \right ) is
m\cdot m_{1}= -1
\Rightarrow m_{1}= \frac{-1}{m}= \frac{-1}{-\sqrt{3}}= \frac{1}{\sqrt{3}}

Let slope of L_{2} be m_{2}

Angle between L_{1} and L_{2}
\tan \left (45 \degree \right )= \left | \frac{\frac{1}{\sqrt{3}}-m_{2}}{1+\frac{1}{\sqrt{3}}m_{2}} \right |
\Rightarrow \left | \frac{1-\sqrt{3}m_{2}}{\sqrt{3}+m_{2}} \right |= 1
\Rightarrow \frac{1-\sqrt{3}m_{2}}{\sqrt{3}+m_{2}} = 1\: or\: \frac{1-\sqrt{3}m_{2}}{\sqrt{3}+m_{2}}= -1
\Rightarrow 1-\sqrt{3}m_{2}= \sqrt{3}+m_{2}\: \: or\: \: 1-\sqrt{3}m_{2}= -\sqrt{3}-m_{2}
\Rightarrow m_{2}= \frac{1-\sqrt{3}}{1+\sqrt{3}}\: or\: m_{2}= \frac{1+\sqrt{3}}{\sqrt{3}-1}
 

Posted by

Kuldeep Maurya

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