English

The coordinates of the point which is equidistant from the three vertices of the ΔAOB as shown in the figure is ______.

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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)`

MCQ
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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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Chapter 7: Coordinate Geometry - Exercise 7.1 [Page 79]

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NCERT Exemplar Mathematics Exemplar [English] Class 10
Chapter 7 Coordinate Geometry
Exercise 7.1 | Q 15 | Page 79

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