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Question
The coordinates of the point which is equidistant from the three vertices of the ΔAOB as shown in the figure is ______.
Options
(x, y)
(y, x)
`(x/2, y/2)`
`(y/2, x/2)`
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Solution
The coordinates of the point which is equidistant from the three vertices of the ∆AOB as shown in the figure is (x, y).
Explanation:
Let the coordinate of the point which is equidistant from the three vertices 0(0, 0), A(0, 2y) and B(2x, 0) is P(h, k).
Then, PO = PA = PB
⇒ (PO)2 = (PA)2 = (PB)2 ...(i)
By distance formula,
`[sqrt((h - 0)^2 + (k - 0)^2)]^2`
= `[sqrt((h - 0)^2 + (k - 2y)^2)]^2`
= `[sqrt((h - 2x)^2 + (k - 0)^2)]^2`
⇒ h2 + k2 = h2 + (k – 2y)2
= (h – 2x)2 + k2 ...(ii)
Taking first two equations, we get
h2 + k2 = h2 + (k – 2y)2
⇒ k2 = k2 + 4y2 – 4yk
⇒ 4y(y – k) = 0
⇒ y = k ...[∵ y ≠ 0]
Taking first and third equations, we get
h2 + k2 = (h – 2x)2 + k2
⇒ h2 = h2 + 4x2 – 4xh
⇒ 4x(x – h) = 0
⇒ x = h ...[∵ x ≠ 0]
∴ Required points = (h, k) = (x, y)
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