# (X, Y) is equidistant from (0,4) , (4,2) and (6,3). What are X and Y?

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### 2 Answers

The distance between any two points (x1, y1) and (x2, y2) is given as: sqrt [(x2- x1) ^2 + (y2- y1) ^2]

We use this to find the distance between (X, Y) and (0, 4). This should be equal to the distance between (X, Y) and (4, 2).

So sqrt [(X-0) ^2 +(Y-4) ^2] = sqrt [(X-4) ^2+(Y-2) ^2]

=> (X-0) ^2 +(Y-4) ^2 = (X-4) ^2+(Y-2) ^2

=> X^2 + Y^2 - 8Y +16 = X^2 - 8X + 16 + Y^2 -4Y +4

=> -8Y = -8X - 4Y + 4.

=> 8X - 8Y + 4Y - 4 = 0

=> 8X - 4Y - 4 = 0

=> 2X - Y -1 = 0 ... (1)

Also we know that the distance between (X, Y) and (0, 4) and the distance between (X, Y) and (6, 3) is equal

=> sqrt [ (X-0)^2+(Y-4)^2 = sqrt[(X-6)^2+(Y-3)^2]

=> X^2 + (Y-4)^2 = (X-6)^2 + (Y-3)^2

=> -8Y + 16 = -12X + 36 -6Y + 9

=> 12X - 8Y + 6Y -29 = 0

=> 12X - 2Y -29 = 0 ... (2)

Now we use (1) and (2) to solve for X and Y

(2) - 2*(1)

=> 8X = 29-2 = 27

=> X = 27/8

Substituting X = 27/8 in 2X - Y -1 = 0

=> Y = 2X-1

=> Y = 2*(27/8)-1

=> Y = 23/4

**Therefore (X, Y) is (27/8, 23/4).**

Since (x,y) is equidistant,from (0,4), (4,2) and (6,3) we get:

The distance between (x, y) and (0,4) = the distance between (x,y) and (4,2):

(x-0)^2 +(y-4)^2 = (x-4)^2+(y-2)^2 or

x^2+ y^2-8y+ 16 = x^2-8x+16 +y^2-4y+4

-8y = -8x-4y+4. or

8x-8y +4y = 4

8x-4y = 4

2x-y = 1...................................(1)

Also the distance between (x,y) and (0,4) and that between (6,3) should be equal:

x^2+(y-4)^2 = (x-6)^2+(y-3)^2

-8y+16 = -12x+36-6y+9

12x-8y+6y = 29

12x-2y = 29.............(2)

2x-y = 1............(1)

(2) - 2(1) gives: 8x = 29-2 = 27. So x = 27/8

From (1) 2x-y = 1. So y = 2x-1 = 2*(27/8)-1 = 23/4

Therefore (x,y) = (27/8 , 23/4) is the point which is equidistant from the given 3 points.