Find the equation of the locus of a point A(x, y) which is equidistant from B(0, 2) and C(2, 1)

1. Analyze the question

We are looking for the equation of the locus of a point $A(x, y)$ that is equidistant from two fixed points, $B(0, 2)$ and $C(2, 1)$. This means the distance from $A$ to $B$ is equal to the distance from $A$ to $C$.

2. Explain the concept

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Since the point $A(x, y)$ is equidistant from $B$ and $C$, then $AB = AC$. To make the calculation easier, we square both sides: $AB^2 = AC^2$

3. Show the solution

Step 1: Set up the distance equations. For $AB^2$: $AB^2 = (x - 0)^2 + (y - 2)^2$ $AB^2 = x^2 + (y^2 - 4y + 4)$

For $AC^2$: $AC^2 = (x - 2)^2 + (y - 1)^2$ $AC^2 = (x^2 - 4x + 4) + (y^2 - 2y + 1)$

Step 2: Equate $AB^2$ and $AC^2$. $x^2 + y^2 - 4y + 4 = x^2 - 4x + 4 + y^2 - 2y + 1$

Step 3: Simplify the equation. Subtract $x^2$ and $y^2$ from both sides: $-4y + 4 = -4x + 4 - 2y + 1$

Combine the constants and variables: $-4y + 4 = -4x - 2y + 5$

Move all terms involving $x$ and $y$ to one side and constants to the other: $4x - 4y + 2y = 5 - 4$ $4x - 2y = 1$

4. Identify the answer

Comparing our result $4x - 2y = 1$ to the given options, it matches option C.

5. Tip

When dealing with "equidistant" problems, always square both sides immediately to get rid of the square root sign. This saves time and prevents errors involving radicals. Also, remember that the locus of a point equidistant from two points is simply the perpendicular bisector of the line segment joining those two points!