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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.
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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