Точка, равноудалённая от A(5,1), B(-3,-7), C(7,-1)

Problem

Find the point that is equidistant from $A(5,1)$, $B(-3,-7)$, and $C(7,-1)$.

Step 1: Write Equal-Distance Equations

Let the unknown point be $P(x,y)$. To avoid square roots, compare squared distances.

Set the squared distance from $P$ to $A$ equal to the squared distance from $P$ to $B$:

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

This simplifies to:

$$x+y+2=0$$

Now set the squared distance from $P$ to $A$ equal to the squared distance from $P$ to $C$:

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

This simplifies to:

$$x-y-6=0$$

Step 2: Solve the Line System

The two equations are:

$$x+y=-2$$

and

$$x-y=6$$

Add the equations to eliminate $y$:

$$2x=4$$

So:

$$x=2$$

Substitute back into $x+y=-2$:

$$2+y=-2$$

Therefore:

$$y=-4$$

So the candidate point is:

$$(2,-4)$$

Step 3: Check the Equal Distances

Check the squared distance from $(2,-4)$ to each point.

To $A(5,1)$:

$$(2-5)^2+(-4-1)^2=(-3)^2+(-5)^2=9+25=34$$

To $B(-3,-7)$:

$$(2+3)^2+(-4+7)^2=5^2+3^2=25+9=34$$

To $C(7,-1)$:

$$(2-7)^2+(-4+1)^2=(-5)^2+(-3)^2=25+9=34$$

Each squared distance is $34$, so the point equidistant from $A$, $B$, and $C$ is:

$$\boxed{(2,-4)}$$