Answer :

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 =


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