Question

Find the coordinates of the point equidistant from three given points A(5,1), B(-3.-7) and
C(7,-1).​

Answer 1

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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

$\sqrt{(x-5)^{2}+ (y-1)^{2}}= \sqrt{(x+3)^{2}+ (y+7)^{2}} = \sqrt{(x - 7)^{2}+ (y+1)^{2}}$

⇒ 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)

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Answer 2

Step-by-step explanation:

given,

the point is equidistant from A(5,1) B(-3,-7) C(7,-1)

let D(x,y) which is equidistant from three points

AD=BD=CD

_________ __________ _________

√(x-5)²+(y-1)²=√(x+3)²+(y+7)²=√(x-7)²+(y+1)²

squaring on all sides

=>(x-5)²+(y-1)²=(x+3)²+(y+7)²=(x-7)²+(y+1)²

=>x²+25-10x+y²+1-2y=x²+9+6x+y²+49+14y=x²+49-14x+y²+1+2y

=>25-10x-2y+1=9+6x+49+14y=49-14x+1+2y

Take first and second equation

=>25-10x-2y+1=9+6x+49+14y

=>25-10x-2y+1-9-6x-49-14y=0

=>-16x-16y-32=0

=>-16(x+y+2)=0

=>x+y+2=0.......................(1)

Take first equation and third equation

=>25-10x-2y+1=49-14x+1+2y

=>25-10x-2y+1-49+14x-1-2y=0

=>4x-4y-24=0

=>4(x-y-6)=0

=>x-y-6=0......................(2)

Add (1) and (2)

=>x+y+2+x-y-6=0

=>2x-4=0

=>x=4/2=2

Substitute x=2 in (1)

=>2+y+2=0

=>y=-4

(x,y)=(2,-4)

HOPE U CAN UNDERSTAND