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  • P1 P2 are points on either of the two lines y square root of 3 x is equal to 2 at a distance of 5 units from their point of intersection Find the coordinates of the foot of perpendiculars drawn from P1 P2 on the bisector of the angle between the given

Q: P1, P2 are points on either of the two lines y-\sqrt{3}\left | x \right |=2 at a distance of 5 units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from P1, P2 on the bisector of the angle between the given lines.

Answers (1)

Given lines are   y-\sqrt{3}\left | x \right |=2

If x≥0, then    y-\sqrt{3} x =2..............(i)

If x<0 then  y+\sqrt{3}x=2.............(ii)

On adding both the equations we get y-\sqrt{3}x +y+\sqrt{3}x=2+2

2y=4  

 y=2

 Putting the value of y=2 in equation (ii), we get  2+\sqrt{3}x=2

\sqrt{3}x=2-2=0

 Point of intersection of given lines is (0,2 )

 Now we find the slopes of given lines  Slope of equation (i) is   y+\sqrt{3}x=2

 Comparing the above equation with y=mx+b, we get  m=\sqrt{3} 

 and we know that m=\tan \theta =\sqrt{3}

 θ= 600

 Slope of equation (ii) is  y=-\sqrt{3}x+2 , we get m=-\sqrt{3}

\tan \theta =-\sqrt{3}

 θ=1800- 600= 1200

In ACB,

\cos 30^{\circ}=\frac{BA}{AC} 

\frac{\sqrt{3}}{2}=\frac{BA}{5}

BA=\frac{5\sqrt{3}}{2}

OB=OA+AB=2+\frac{5\sqrt{3}}{2}

Hence, the coordinates of the foot of perpendicular =\left (0,2+\frac{5\sqrt{3}}{2} \right )

 

 

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