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

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by yashRavi

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by yashRavi

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The only point which will be equidistant from the three given points is the circum-center O of the triangle formed by the points A (5, 3), B(5, -5) and C (1, -5).

It lies on the intersection of the perpendicular bisectors of the sides AB, BC and CA.

Mid point of AB = D = [ (5+5)/2, (3-5)/2 ] = (5, -1)

Mid point of BC = E = [ (5+1)/2 , (-5-5)/2 ] = (3, -5)

slope of AB = (-5 -3)/ (5-5) = ∞

=> Slope of perpendicular bisector DO of AB = 0

equation of DO : y = 0 *x + c

as D lies on it, -1 = c =>* DO is : y = -1.*

slope of BC = (-5 +5) / (1 -5) = 0

=> slope of the perpendicular bisector EO of BC is = ∞.

=> equation of EO is : x = c

as E lies on EO, 3 = c.

=>* EO is x = 3. *

The intersection of DO and EO is the circumcenter O : (3, -1).

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another way:

let the point equidistant be O (x, y).

Hence, OA² = OB² = OC²

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

From the first equality, we get:

y - 3 = - (y + 5) as y - 3 ≠ y + 5

2 y = -2 => y = -1

From the second equality we get :

x - 5 = - (x - 1) as x - 5 ≠ x - 1

2 x = 6 => x = 3

So (3 , -1) is the equidistant point.

It lies on the intersection of the perpendicular bisectors of the sides AB, BC and CA.

Mid point of AB = D = [ (5+5)/2, (3-5)/2 ] = (5, -1)

Mid point of BC = E = [ (5+1)/2 , (-5-5)/2 ] = (3, -5)

slope of AB = (-5 -3)/ (5-5) = ∞

=> Slope of perpendicular bisector DO of AB = 0

equation of DO : y = 0 *x + c

as D lies on it, -1 = c =>

slope of BC = (-5 +5) / (1 -5) = 0

=> slope of the perpendicular bisector EO of BC is = ∞.

=> equation of EO is : x = c

as E lies on EO, 3 = c.

=>

The intersection of DO and EO is the circumcenter O : (3, -1).

=============================

another way:

let the point equidistant be O (x, y).

Hence, OA² = OB² = OC²

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

From the first equality, we get:

y - 3 = - (y + 5) as y - 3 ≠ y + 5

2 y = -2 => y = -1

From the second equality we get :

x - 5 = - (x - 1) as x - 5 ≠ x - 1

2 x = 6 => x = 3

So (3 , -1) is the equidistant point.