Math Problem Statement

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

Solution

To find the equation of the locus of point PP that is equidistant from the points (2,3)(-2, 3) and (4,1)(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)(-2, 3) and (4,1)(4, -1).

The midpoint MM is calculated as follows:

M=(2+42,3+(1)2)=(22,22)=(1,1)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)(-2, 3) and (4,1)(4, -1):

The slope mm is given by:

m=y2y1x2x1=134(2)=46=23m = \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 mbisectorm_{\text{bisector}} is:

mbisector=32m_{\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:

yy1=m(xx1)y - y_1 = m(x - x_1)

Using the midpoint (1,1)(1, 1) and the slope 32\frac{3}{2}:

y1=32(x1)y - 1 = \frac{3}{2}(x - 1)

Simplify and rearrange to get the equation in standard form:

y1=32(x1)y - 1 = \frac{3}{2}(x - 1) y1=32x32y - 1 = \frac{3}{2}x - \frac{3}{2} 2(y1)=3(x1)2(y - 1) = 3(x - 1) 2y2=3x32y - 2 = 3x - 3 3x2y=13x - 2y = 1

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

3x2y=13x - 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.

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Perpendicular Bisector

Formulas

Midpoint Formula
Point-Slope Form
Distance Formula

Theorems

Perpendicular Bisector Theorem

Suitable Grade Level

Grades 9-12