Problem 13 - Contemporary Precalculus

Question

Find the distance between the two points and the midpoint of the segment joining them. $$(-3,5),(2,-7)$$

Step-by-Step Solution

Answer

Answer: The distance between the points (-3,5) and (2,-7) is 13, and the midpoint of the line segment joining them is (-1/2, -1).

Step 1: Identify the given points

The two given points are \((-3,5)\) and \((2,-7)\).

Step 2: Use the distance formula to find the distance between the points

The distance formula is given by $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ Plug in the given points \((-3,5)\) and \((2,-7)\), where \((x_1, y_1) = (-3,5)\) and \((x_2, y_2) = (2,-7)\). Calculate the distance: $$d = \sqrt{(2 - (-3))^2 + (-7 - 5)^2}$$

Step 3: Simplify and solve for d

Simplify the expression inside the square root: $$d = \sqrt{(2 + 3)^2 + (-7 - 5)^2} = \sqrt{5^2 + (-12)^2}$$ Now, simplify further: $$d = \sqrt{25 + 144} = \sqrt{169}$$ Thus, the distance between the two points is \(d = 13\).

Step 4: Use the midpoint formula to find the midpoint of the segment

The midpoint formula is given by $$M(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$$ Plug in the given points \((-3,5)\) and \((2,-7)\) where \((x_1, y_1) = (-3,5)\) and \((x_2, y_2) = (2,-7)\). Calculate the midpoint: $$M(\frac{-3+2}{2}, \frac{5-7}{2})$$

Step 5: Simplify and find the midpoint

Simplify the expression inside the parentheses: $$M(\frac{-1}{2}, \frac{-2}{2})$$ Thus, the midpoint of the line segment joining the given points is \(M(-\frac{1}{2}, -1)\). To summarize, the distance between the two points \((-3,5)\) and \((2,-7)\) is 13, and the midpoint of the line segment joining them is \((-\frac{1}{2}, -1)\).

Key Concepts

Distance Formula

Understanding how to measure the distance between two points on a coordinate plane is essential in analytical geometry. The Distance Formula allows you to calculate this distance effectively. It is derived from the Pythagorean Theorem and is used to find the straight line distance between two given points.

This formula is:

• \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Let's break this down:

• The difference \((x_2 - x_1)\) represents the horizontal change between two points on the x-axis.
• The difference \((y_2 - y_1)\) represents the vertical change between the same points on the y-axis.

For example, if you have the points \((-3,5)\) and \(2,-7)\), substituting into the Distance Formula yields the calculation of:

• \[d = \sqrt{(2 - (-3))^2 + (-7 - 5)^2} = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13\]

By understanding each part of the formula, you can accurately determine the distance between any two points in coordinate geometry.

Midpoint Formula

To find the midpoint of a line segment joining two points, use the Midpoint Formula. This formula gives you the average of the x-coordinates and the y-coordinates of the given points. The result provides the exact center point of the segment.

Here’s the formula:

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

Breaking it down:

• The expression \((x_1+x_2)/2\) calculates the average of the x-values.
• The expression \((y_1+y_2)/2\) calculates the average of the y-values.

Using the given points of \((-3,5)\) and \(2,-7)\), it follows:

• \[M\left(\frac{-3+2}{2}, \frac{5-7}{2}\right) = M\left(\frac{-1}{2}, \frac{-2}{2}\right) = M\left(-\frac{1}{2}, -1\right)\]

Thus, the midpoint formula helps efficiently find the middle of any line segment described by two points in the coordinate plane.

Coordinate Geometry

Coordinate Geometry, sometimes called Analytic Geometry, is the study where you apply algebraic principles to geometric problems. This field is essential as it connects shapes with algebra, allowing for deeper analysis and understanding of geometric properties.

Key applications include:

• Finding distances and midpoints, which help in determining relative positions on the Cartesian plane.
• Equation of a line, used to describe linear paths or boundaries.
• Slopes and intersections, which determine angles and meeting points between lines.

The Cartesian coordinate system forms the backbone of this study, enabling the description of each point with an ordered pair \((x, y)\). This framework makes other calculations like finding equations and analyzing geometrical figures geometrics more straightforward as they transition into algebraic work.

By integrating algebra with geometry, Coordinate Geometry provides a powerful tool for solving complex problems involving shapes, positions, and dimensions effectively.