a moving point p is always equidistant from the points (-2, 3) and (4, -1). Find the equation of the point p

To find the equation of the locus of point \( P \) that is equidistant from the points \((-2, 3)\) and \( (4, -1)\), we use the concept that the locus of all points equidistant from two given points is the perpendicular bisector of the line segment joining the two points.

First, let's determine the midpoint of the segment connecting \((-2, 3)\) and \( (4, -1)\).

The midpoint \( M \) is calculated as follows:

\[ M = \left( \frac{-2 + 4}{2}, \frac{3 + (-1)}{2} \right) = \left( \frac{2}{2}, \frac{2}{2} \right) = (1, 1) \]

Next, we need to find the slope of the line passing through \((-2, 3)\) and \( (4, -1)\):

The slope \( m \) is given by:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 3}{4 - (-2)} = \frac{-4}{6} = -\frac{2}{3} \]

The slope of the perpendicular bisector is the negative reciprocal of this slope. Thus, the slope of the perpendicular bisector \( m_{\text{bisector}} \) is:

\[ m_{\text{bisector}} = \frac{3}{2} \]

Now, we use the point-slope form of the line equation to write the equation of the perpendicular bisector. The point-slope form is:

\[ y - y_1 = m(x - x_1) \]

Using the midpoint \( (1, 1) \) and the slope \( \frac{3}{2} \):

\[ y - 1 = \frac{3}{2}(x - 1) \]

Simplify and rearrange to get the equation in standard form:

\[ y - 1 = \frac{3}{2}(x - 1) \]
\[ y - 1 = \frac{3}{2}x - \frac{3}{2} \]
\[ 2(y - 1) = 3(x - 1) \]
\[ 2y - 2 = 3x - 3 \]
\[ 3x - 2y = 1 \]

Thus, the equation of the locus of the point \( P \), which is equidistant from \((-2, 3)\) and \( (4, -1)\), is:

\[ 3x - 2y = 1 \]

Do you want details or have any questions?

Here are 8 related questions that expand on this information:
1. What is the general form of the equation of a line?
2. How do you find the distance between two points in the coordinate plane?
3. What is the significance of the midpoint in geometry?
4. How do you determine the slope of a line given two points?
5. What is the relationship between the slopes of perpendicular lines?
6. How can you convert the point-slope form of a line equation to the standard form?
7. What are the applications of the perpendicular bisector in real-life problems?
8. How do you verify if a point lies on a given line?

**Tip:** Understanding the properties of perpendicular bisectors can greatly simplify problems involving equidistant points.

Learn how to find the equation of a point equidistant from the points (-2, 3) and (4, -1) using coordinate geometry. This problem involves calculating the midpoint, determining slopes, and applying the perpendicular bisector theorem. Suitable for high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation