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P is point on either of the two lines y-\sqrt{3}\left | x \right |=2 at a distance of 5 units from their point of intersection. The co-ordinates of the foot of the perpendicular from P on the bisector of the angle between them are :-

Option: 1

\left ( 0,\frac{4+5\sqrt{3}}{2} \right ) or \left ( 0,\frac{4-5\sqrt{3}}{2} \right )depending on which the points p is taken
 


Option: 2

\left ( 0,\frac{4+5\sqrt{3}}{2} \right )


Option: 3

\left ( 0,\frac{4-5\sqrt{3}}{2} \right )

 


Option: 4

\left ( \frac{5}{2},\frac{5\sqrt{3}}{2} \right )


Answers (1)

 

Distance formula -

The distance between the point A\left ( x_{1},y_{1} \right )\: and \: B\left ( x_{2},y_{2} \right )

is \sqrt{\left ( x_{1} -x_{2}\right )^{2}+\left ( y_{1} -y_{2}\right )^{2}}

- wherein

 

 

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

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

P=\left ( \frac{5}{2},\frac{4+5\sqrt{3}}{2} \right ) or \left (- \frac{5}{2},\frac{4+5\sqrt{3}}{2} \right )

distance of p on its angle bisector i.e. y-axis is \left ( 0,\frac{4+5\sqrt{3}}{2} \right )

Posted by

Sumit Saini

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