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

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).