Introduction to the Locus of a Point

In geometry, the locus of a point refers to the set of all points that satisfy a given condition. In this article, we will explore the specific case where a point is equidistant from two given points, A (3, -2) and B (2, 1). We will derive the equation of the locus, which is the perpendicular bisector of the line segment connecting these two points.

Understanding the Perpendicular Bisector

The locus of points that are equidistant from two given points A and B is the perpendicular bisector of the line segment AB. This means that any point on this line is equally distant from A and B.

Calculating the Perpendicular Bisector

Find the Midpoint: The first step is to find the midpoint of the line segment AB. Calculate the Midpoint: The formula for the midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is:

M (left(frac{x_1 x_2}{2}, frac{y_1 y_2}{2}right))

Using points A (3, -2) and B (2, 1), the midpoint is:

M  (left(frac{3   2}{2}, frac{-2   1}{2}right)  left(frac{5}{2}, -frac{1}{2}right))

Find the Slope of AB: The slope of a line segment with endpoints (x1, y1) and (x2, y2) is given by:

Slope of AB (frac{y_2 - y_1}{x_2 - x_1})

For points A (3, -2) and B (2, 1), the slope is:

Slope of AB  (frac{1 - (-2)}{2 - 3}  frac{3}{-1}  -3)

Find the Slope of the Perpendicular Bisector: The slope of the perpendicular bisector is the negative reciprocal of the slope of AB. Calculate the Equation of the Perpendicular Bisector: Using the point-slope form of the line equation (y - y_1 m(x - x_1)), where (x1, y1) is the midpoint and m is the slope, we get: Substitute the Values: Here, the midpoint (x1, y1) is (left(frac{5}{2}, -frac{1}{2}right)) and the slope (m frac{1}{3}). Derive the Final Equation: The equation is:

y   frac{1}{2}  frac{1}{3}left(x - frac{5}{2}right)

Simplifying this, we get:

y  frac{1}{3}x - frac{5}{6} - frac{3}{6}  frac{1}{3}x - frac{4}{3})

Conclusion

The locus of all points that are equidistant from points A (3, -2) and B (2, 1) is given by the equation:

y  frac{1}{3}x - frac{4}{3}

This line represents the perpendicular bisector of the segment AB. Any point on this line is equidistant from points A and B.

Additional Analysis

As an alternative, one can consider the points equidistant from A (3, -2) and B (2, 1) using the distance formula. Let P (x, y) be any point such that PA PB.

Using the distance formula,

sqrt{(x - 3)^2   (y   2)^2}  sqrt{(x - 2)^2   (y - 1)^2}

Squaring both sides gives:

(x - 3)^2   (y   2)^2  (x - 2)^2   (y - 1)^2

Expanding and simplifying this equation results in:

x^2 - 6x   9   y^2   4y   4  x^2 - 4x   4   y^2 - 2y   1

Cancelling like terms, we get:

-2x   6y   8  0 text{ or } x - 3y - 4  0

This is the equation of a straight line which is also the locus of points equidistant from A and B.