P₁, P₂ are points on either of the two lines 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.

Given lines are y - √3|x| = 2

If x ≥ 0, then

y - √3x = 2 …(i)

If x < 0, then

y + √3x = 2 …(ii)

On adding eq. (i) and (ii), we get

y - √3x + y + √3x = 2 + 2

⇒ 2y = 4

⇒ y = 2

Putting the value of y = 2 in eq. (ii), we get

2 + √3x = 2

⇒ √3x = 2 – 2

⇒ x = 0

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

Now, we find the slopes of given lines.

Slope of eq. (i) is

y = √3x + 2

Comparing the above equation with y = mx + b, we get

m = √3

and we know that, m = tan θ

∴ tan θ = √3

⇒ θ = 60° [∵ tan 60° = √3]

Slope of eq. (ii) is

y = - √3x + 2

Comparing the above equation with y = mx + b, we get

m = -√3

and we know that, m = tan θ

∴ tan θ = -√3

⇒ θ = (180° - 60°)

⇒ θ = 120°

In ΔACB,

[given: AC = 5units]

∴ OB = OA + AB

Hence, the coordinates of the foot of perpendicular =