# Find the coordinates of the point equidistant from point A(1, 2), B(3, -4), and C(5, -6)?

We will be using the distance formula for finding the coordinates of the point.

## Answer: The coordinates of the point equidistant from point A(1, 2), B(3, -4), and C(5, -6) are P(11, 2).

Let's get the solution step by step.

**Explanation**:

The three given points are A(1, 2), B(3, -4), and C(5, -6).

Let P (x, y) be the point that is equidistant from the given three points.

So, PA = PB = PC ---- (1)

According to distance formula, we have distance between any two points (x, y) and (\(x_{1}\), \(y_{1}\)) = √(x - \(x_{1}\))^{2 }+ (y - \(y_{1}\))^{2}.

Distance PA = √(x - 1)^{2 }+ (y - 2)^{2}

Distance PB = √(x - 3)^{2 }+ (y + 4)^{2}

Distance PC = √(x - 5)^{2 }+ (y + 6)^{2}

Substituting these in equation (1),

√(x - 1)^{2 }+ (y - 2)^{2} = √(x - 3)^{2 }+ (y + 4)^{2} = √(x - 5)^{2 }+ (y + 6)^{2}

⇒ x^{2} + 1 - 2x + y^{2} + 4 – 4y = x^{2} + 9 - 6x + y^{2} + 16 + 8y = x^{2} + 25 – 10x + y^{2} + 36 + 12y

⇒ 1 – 2x + 4 – 4y = 9 - 6x + 16 + 8y = 25 – 10x + 36 + 12y

Considering, 1 – 2x + 4 – 4y = 9 - 6x + 16 + 8y, we get,

⇒ 2x - 6y = 10 … (1)

Now, consider, 1 – 2x + 4 – 4y = 25 – 10x + 36 + 12y

⇒ x – 2y = 7 … (2)

Solving (1) and (2), we get x = 11, y = 2.

You can solve the system of equations using Cuemath's linear equation calculator.

Thus, the required point is (11, 2).

We can verify our answer by using the distance algebra calculator.

### Hence, the coordinates of the point equidistant from point A(1, 2), B(3, -4), and C(5, -6) are P(11, 2).

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