Problem 52 - Calculus: One and Several Variables

Question

The perpendicular bisector of the line scement \(\overline{P Q}\) is the line which is perpendicular to \(\bar{P}(2\) and passes through the midpoint of \(\overline{P Q}\). Find an equation for the perpendicular bisector of the line segment that joins the two points. $$P(-1,3), \quad Q(3,-4)$$

Step-by-Step Solution

Answer

The equation of the perpendicular bisector of the line segment that joins points P(-1,3) and Q(3,-4) is \(y=\frac{4}{7}x-\frac{7}{2}\).

Step 1: Finding the Midpoint

By definition, the midpoint of a line segment \(\overline{P Q}\) with endpoints \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) is \(\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)\). Substituting \(P(-1,3)\) and \(Q(3,-4)\) into the midpoint formula gives \(M(\frac{-1+3}{2}, \frac{3-4}{2}) = (1, -0.5)\). So, the midpoint M is (1, -0.5)

Step 2: Calculate the Perpendicular Slope

The slope of the line segment \(\overline{P Q}\) is given by \(m = \frac{y_2-y_1}{x_2-x_1}\). Substituting the given points into the formula gives \(m = \frac{-4-3}{3-(-1)} = -\frac{7}{4}\). The slope of the line perpendicular to \(\overline{P Q}\) is the negative reciprocal of \(m\) which is \(\frac{4}{7}\)

Step 3: Finding the Equation of the Perpendicular Bisector

The equation of a line in slope-intercept form is given by \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept. Here m, the slope of the perpendicular bisector, is found to be \(\frac{4}{7}\). We can find \(c\) by substituting the x and y coordinates of the midpoint M into the equation. This gives \(-0.5=\frac{4}{7} * 1 + c\), which solves for \(c = -\frac{7}{2}\). Thus, the equation of the perpendicular bisector of \(\overline{P Q}\) is \(y=\frac{4}{7}x-\frac{7}{2}\)

Key Concepts

Understanding the Midpoint Formula

The **midpoint formula** is a valuable tool when you need to find the center point between two locations on a graph. If you have a line segment with endpoints \(P(x_1, y_1)\) and \(Q(x_2, y_2)\), the midpoint \(M\) is found by averaging the x-coordinates together and the y-coordinates together. It’s expressed as:

• \(M = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)\)

Let's break this down: add the x-values from both points, divide by 2, and then do the same for the y-values. This gives you the point that lies exactly in the middle of \(P\) and \(Q\).
For example, using the points \((-1, 3)\) and \((3, -4)\), the midpoint \(M\) is calculated as:

• \(x_1 + x_2 = -1 + 3 = 2 \Rightarrow \frac{2}{2} = 1\)
• \(y_1 + y_2 = 3 - 4 = -1 \Rightarrow \frac{-1}{2} = -0.5\)

Thus, the midpoint is \((1, -0.5)\), which will be crucial for constructing the perpendicular bisector of the line segment.

Calculating the Slope of a Line

The **slope of a line** represents how steep the line is and is often denoted by \(m\). It's calculated by taking the difference in the y-coordinates and dividing by the difference in the x-coordinates of two points on the line. Mathematically, this is expressed as:

• \(m = \frac{y_2 - y_1}{x_2 - x_1}\)

For example, if you have two points, \(P(-1, 3)\) and \(Q(3, -4)\), plug these into the formula:

• \(m = \frac{-4 - 3}{3 - (-1)} = \frac{-7}{4}\)

A key point to note here is the concept of the **perpendicular slope**. If two lines are perpendicular, the slope of one is the negative reciprocal of the other. Hence, the slope of the perpendicular line to \(PQ\) becomes:

• \(-m = \frac{4}{7}\)

This perpendicular slope is essential for finding the perpendicular bisector of a line segment.

Finding the Equation of a Line

To find the **equation of a line**, we often employ the slope-intercept formula which is \(y = mx + c\), where:

• \(m\) is the slope.
• \(c\) is the y-intercept, indicating where the line crosses the y-axis.

For the perpendicular bisector, we already know the slope is \(\frac{4}{7}\). We can substitute the midpoint \((1, -0.5)\) into the line equation to find \(c\):

• \(-0.5 = \frac{4}{7} \cdot 1 + c\)
• Simplifying this, we solve for \(c = -\frac{7}{2}\).

Thus, the equation for the perpendicular bisector of the line segment joining \(P\) and \(Q\) is:

• \(y = \frac{4}{7}x - \frac{7}{2}\)

This line is normal to \(PQ\) and passes through the midpoint, cutting the line segment into two equal parts at a right angle.