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Question
P1, P2 are points on either of the two lines `- sqrt(3) |x|` = 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.
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Solution
Given lines are `- sqrt(3) |x|` = 2
⇒ `y - sqrt(3)x` = 2, if x ≥ 0 .....(i)
And `y + sqrt(3)x` = 2, if x < 0 ......(ii)
Slope of equation (i) is tan θ = `sqrt(3)`
∴ θ = 60°
Slope of equation (ii) is tan q `- sqrt(3)`
∴ θ = 120°
Solving equation (i) and equation (ii) we get
`y - sqrt(3) = 2`
`y + sqrt(3)x = 2`
2y = 4
⇒ y = 2
Putting the value of y is eq. (i) we get
x = 0
∴ Point of intersection of line (i) and (ii) is Q(0, 2)
∴ QO = 2
In ΔPEQ,
cos 30° = `"PQ"/"QE"`
`sqrt(3)/2 = "PQ"/5`
∴ PQ = `(5sqrt(3))/2`
∴ OP = OQ + PQ
= `2 + (5sqrt(3))/2`
Hence, the coordinates of the foot of perpendicular = `(0, 2 + (5sqrt(3))/02)`.
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