find the coordinates of the point equidistant from three given points A (5 , 1 ) , B (-3,-7 ) and C( 7, -1).
Hi!
Here is the answer to your question.
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)
Hope! This helps you.
Cheers!