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

⇒

*x*^{2}+ 25 – 10*x*+*y*^{2}+ 1 – 2*y*=*x*^{2}+ 9 + 6*x*+*y*^{2}+ 49 + 14*y*=*x*^{2}+ 49 – 14*x*+*y*^{2}+ 1 + 2*y*⇒ 25 – 10

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

*x*+ 1 – 2*y*= 9 + 6*x*+ 49 + 14*y*⇒ 26 = 16

*x*+ 58 + 16*y*⇒ 16

*x*+ 32 + 16*y*= 0⇒

*x*+ 2 +*y*= 0 … (1)25 – 10

*x*+ 1 – 2*y*= 49 – 14*x*+ 1 + 2*y*⇒ 26 + 4

*x*– 4*y*= 50⇒ 4

*x*– 4*y*= 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!

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