Problem 32 - Algebra 1: Concepts and Skills

Question

Find the midpoint of the line segment connecting the given points. Then show that the midpoint is the same distance from each point. \((-4,-2),(10,-6)\)

Step-by-Step Solution

Answer

The midpoint of the line segment connecting the points \((-4,-2)\) and \((10,-6)\) is \((3,-4)\), and this point is \(\sqrt{53}\) units distant from both given points.

Step 1: Find the Midpoint

We can find the midpoint of a line segment with the given endpoints \(A(-4,-2)\) and \(B(10,-6)\) by averaging the x-coordinates and the y-coordinates separately. This can be achieved using the formula: \[ M = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \] Substitute the given points into the formula: \[ M = \left( \frac{-4+10}{2}, \frac{-2+(-6)}{2} \right) = (3,-4) \]

Step 2: Verify the Distances

Now, we have to show that the midpoint is the same distance from each point. The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) can be calculated using the formula: \[ d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \] Calculate the distance from \(A\) to \(M\) and from \(B\) to \(M\): \[ d_{AM} = \sqrt{(3+4)^2+((-4)-(-2))^2} = \sqrt{49+4} = \sqrt{53}\] \[ d_{BM} = \sqrt{(10-3)^2+((-6)-(-4))^2} = \sqrt{49+4} = \sqrt{53} \] Since \(d_{AM} = d_{BM}\), we have shown that the midpoint is the same distance from each point.

Key Concepts

Midpoint Formula

In coordinate geometry, finding the midpoint of a line segment is a fundamental concept that students often encounter. The midpoint of a line segment is the point that divides the segment into two equal parts.

The midpoint formula is a standard algebraic expression used for this purpose. It is given by the coordinates \( M = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \), where \(x_1, y_1\) and \(x_2, y_2\) are the coordinates of the endpoints of the line segment. To find the midpoint, you simply calculate the average of the x-coordinates and the average of the y-coordinates of the endpoints.

Using the formula, we can easily determine the midpoint for any line segment lying on a two-dimensional plane. For instance, for the points \(A(-4,-2)\) and \(B(10,-6)\) given in the original exercise, the midpoint calculation yields \(M = \left( \frac{-4+10}{2}, \frac{-2+(-6)}{2} \right) = (3, -4)\). This straightforward process is essential for students to master, as it is a building block for more complex geometric concepts.

Distance Formula

The distance formula is integral to coordinate geometry, often used to find the length of a line segment among other applications. It is derived from the Pythagorean theorem, which relates the sides of a right triangle to its hypotenuse.

The formula \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \) calculates the distance \(d\) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) in the cartesian plane. In essence, this is finding the hypotenuse of a right triangle whose legs are the differences between the respective x-coordinates and y-coordinates of the two points.

Picturing this geometrically, if you have two points on a grid, the distance formula provides the straight-line distance between them. In the context of the given exercise, it helps validate that the calculated midpoint \(M(3, -4)\) is equidistant from the points \(A(-4, -2)\) and \(B(10, -6)\). The distances from \(A\) and \(B\) to the midpoint both compute to \(\sqrt{53}\), thus confirming the midpoint's position through this proof.

Coordinate Geometry

At its core, coordinate geometry, often known as analytic geometry, is the study of geometry using a coordinate system. This system facilitates the process of defining geometric shapes in a numerical way and enables algebraic equations to represent geometric figures.

This field of mathematics combines algebra and geometry to solve problems involving shapes, sizes, and the relative position of figures. By working with defined axes, typically the x and y-axes in two dimensions, students learn to describe locations and shapes using coordinates and equations.

Key concepts within coordinate geometry include points, line segments, midpoints, distances, and slopes, to name a few. Understanding how to apply the midpoint and distance formulas is crucial for students to analyze geometric relationships and solve related problems. The discipline's incorporation of visual aids (graphs) alongside algebraic calculations empowers learners to interpret and handle complex spatial information efficiently.