The coordinates of the point which is equidistant from the three vertices of the ∆ AOB as shown in the Fig. 7.1 is
a. (x, y)
b. (y, x)
c. x/2, y/2
d. y/2, x/2

Solution:

Consider the coordinate of the point which is equidistant from the three vertices O (0, 0), A (0, 2y) and B (2x, 0) as P (h, k)

Here PO = PA = PB

PO² = PA² = PB² --- (1)

\([\sqrt{(h-0)^{2}+(k-0)^{2}}]^{2}=[\sqrt{(h-0)^{2}+(k-2y)^{2}}]^{2}=[\sqrt{(h-2x)^{2}(k-0)^{2}}]^{2}\)

Now we get

h² + k² = h² + (k - 2y)² = (h - 2x)² + k² --- (1)

Let us consider the first two equation

h² + k² = h² + (k - 2y)²

k² = k² + 4y² - 4yk

Taking out the common terms

4y(y - k) = 0

y = k where y ≠ 0

Considering the first and third equation

h² + k² = (h - 2x)² + k²

h² = h² + 4x² - 4xh

Taking out the common terms

4x(x - h) = 0

x = h where x ≠ 0

So the required points are (h, k) = (x, y)

Therefore, the coordinates of the point are (x, y).

✦ Try This: The perpendicular bisector of the line segment joining the points P (1, 2) and Q (3, 4) cuts the y-axis at

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7

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NCERT Exemplar Class 10 Maths Exercise 7.1 Problem 15

The coordinates of the point which is equidistant from the three vertices of the ∆ AOB as shown in the Fig. 7.1 is a. (x, y), b. (y, x), c. x/2, y/2, d. y/2, x/2

Summary:

The coordinates of the point which is equidistant from the three vertices of the ∆ AOB as shown in the Fig. 7.1 is (x, y)

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