Question

Find the
coordinates of the point
equidistance from the point A(5,1)
B (3,7), C(7,-1)​

Answer 1

Answer:

The given three points are A (5, 1) , B (–3, –7 ) and C (7, –1)

Let P (x, y) be the point equidistant from these three points.

So, PA = PB = PC

⇒ x2 + 25 – 10x + y2 + 1 – 2y = x2 + 9 + 6x + y2 + 49 + 14y = x2 + 49 – 14x + y2 + 1 + 2y

⇒ 25 – 10x + 1 – 2y = 9 + 6x + 49 + 14y = 49 – 14x + 1 + 2y

25 – 10x + 1 – 2y = 9 + 6x + 49 + 14y

⇒ 26 = 16x + 58 + 16y

⇒ 16x + 32 + 16y = 0

⇒ x + 2 + y = 0 … (1)

25 – 10x + 1 – 2y = 49 – 14x + 1 + 2y

⇒ 26 + 4x – 4y = 50

⇒ 4x – 4y = 24

⇒ x – y = 6 … (2)

Solving (1) and (2)

x = 2, y = –4

Thus, the required point is (2, –4)