Problem 34 - Precalculus Mathematics for Calculus

Question

Which of the points \(C(-6,3)\) or \(D(3,0)\) is closer to the point \(E(-2,1) ?\)

Step-by-Step Solution

Answer

Point \(C(-6,3)\) is closer to point \(E(-2,1)\).

Step 1: Understand the Distance Formula

To solve this problem, we need to calculate the distance from points \(C\) and \(D\) to point \(E\). The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) in a plane can be calculated using the distance formula: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]This formula will calculate the Euclidean distance between any two given points.

Step 2: Calculate Distance from C to E

Now, we'll use the distance formula to find the distance between points \(C(-6,3)\) and \(E(-2,1)\):\[d_{CE} = \sqrt{((-2) - (-6))^2 + (1 - 3)^2} = \sqrt{(4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20}\]Therefore, the distance from \(C\) to \(E\) is \(\sqrt{20}\).

Step 3: Calculate Distance from D to E

Next, we calculate the distance between \(D(3,0)\) and \(E(-2,1)\):\[d_{DE} = \sqrt{((-2) - 3)^2 + (1 - 0)^2} = \sqrt{(-5)^2 + 1^2} = \sqrt{25 + 1} = \sqrt{26}\]Thus, the distance from \(D\) to \(E\) is \(\sqrt{26}\).

Step 4: Compare the Distances

To determine which point is closer, compare the two distances:- Distance from \(C\) to \(E\) is \(\sqrt{20}\).- Distance from \(D\) to \(E\) is \(\sqrt{26}\). Since \(\sqrt{20} \approx 4.47\) and \(\sqrt{26} \approx 5.10\), \(\sqrt{20}\) is smaller than \(\sqrt{26}\), so point \(C\) is closer to point \(E\).

Key Concepts

Euclidean distance

The Euclidean distance is a fundamental concept in geometry. It measures the straight-line distance between two points in a plane or in space. Think of it as measuring the direct route from one point to another, just like the crow flies. It's one of the most common ways to calculate distance, especially on a flat surface.

The formula for Euclidean distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is provided by:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]- \(d\) represents the distance,- \(x_1\) and \(y_1\) are the coordinates of the first point,- \(x_2\) and \(y_2\) are the coordinates of the second point.You square the differences between the corresponding coordinates, add them up, and then take the square root of the result. This way, the Euclidean distance accounts for differences in both the horizontal and vertical distances between the points.

compare distances

When faced with a problem that involves finding which of multiple points is closer, we use the Euclidean distance formula to compare distances between each point and a reference point. This comparison involves calculating the distance from each known point to the key point and seeing which distance is shorter.

Here's how you can effectively perform this comparison:

• Calculate the Euclidean distance from the first point to the reference point.
• Calculate the Euclidean distance from the second (or additional) point to the reference point.
• Compare these distances by evaluating the numerical values.
• The point with the smallest distance value is closest to the reference point.

This method is applicable in various practical situations, such as determining the nearest location, selecting the shortest route, or analyzing spatial data.

distance between points

Understanding how to compute the distance between points is essential for solving various geometrical and real-world problems. In coordinate geometry, the position of each point is defined by coordinates, which are numbers that determine the location of the point on a plane.

For instance, if point \(C\) is represented by \((-6, 3)\) and point \(E\) by \((-2, 1)\), you can use these coordinates to find how far apart they are. The Euclidean distance formula aids in converting spatial coordinates into a measurable distance.

• First, substitute the coordinates into the distance formula.
• Then, perform operations such as subtraction, squaring, addition, and square rooting.
• The result gives you the exact distance between the two points.

Calculating such distances helps in various fields including physics, engineering, and cartography, where precision in measuring space is critical.