Question

find the coordinates of the point which is eqidistant from the three vertices of the ∆AOB as shown in the fig.

Answer 1

Mid point of of the Hypotonues is equidistant from the points in a right angles triangle
Therefore by mid point formula
(X, Y) is answer.

Answer 2

Answer:

Step-by-step explanation:

Solution :-

From the given figure, coordinates of three vertices of a triangle, O(0, 0), A(0, 2Y) and B(h, k)

Suppose the required point be P whose coordinates are (h, k)

Now, P is equipment from the three vertices of DAOB,

Therefore,

PO = PA = PB

or, (PO)² = (PA)² = (PB)²

By using distance formula, we get

⇒ [√(h - 0)² + (k - 0)²]² = [√(h - 0)² + (k - 2y)²]² = [√(h - 2x)² + (5 - 0)]²

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

By taking 1st and 2nd parts, we get

⇒ h² + k² = h² + (k - 2x)²

⇒ k² = k² + 4y² - 4yk

⇒ 0 = 4y² - 4yk

⇒ 4y(y - k) = 0

⇒ y = k

By taking 1st and 3rd parts, we get

⇒ h² + k² = (h - 2x)² + k²

⇒ h² = h² + 2x² - 4xh

⇒ 4x(x - h) = 0

⇒ h = x

Hence, the coordinates are (h, k) = (x, y)