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

Hi!

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!

• 126

(2,-4)

• -7

let the points be P(X,Y)

thrfor  pa2=pb2=pc2

solve pa2 ,pb2 and pc2

x+Y=-2

X=-2-Y

SUBSTITUTE X=-2-Y IN ONE OF D EQN...

-8Y=32

Y=-32/8

Y= -8

THRFORE X=2 Y=-4

• 3

Let the coordinates be A(5,1), B(3,-7) and C(7,-1) and the required point be P(x,y). since it is equidistant:

Distance between :  AP=CP=BP

AP2=(x-5)2+(y-1)2

BP2=( x+3)2+(y+7)2

CP2=(x-7)2+(y+1)2

We have AP2=BP2

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

solving:

Or,  -x-y=2

Or,  x=2-y …………..(i)

Also: AP2=Cp2

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

Solving: x-y=6

Or, x=6+y …………………..(ii)

Equating (i) and (ii): 2-y=6+y

Y=-2 and substituting the value in ii.

X=6-2=4

Thus the required point is (4,-2).