Problem 40
Question
Which of the points \(C(-6,3)\) and \(D(3,0)\) is closer to the point \(E(-2,1) ?\)
Step-by-Step Solution
Verified Answer
Point C(-6,3) is closer to E(-2,1) than point D(3,0).
1Step 1: Calculate Distance from C to E
To find which point is closer to point E, we first need to calculate the distance from point C(-6, 3) to point E(-2, 1) using the distance formula. The distance formula between two points \(x_1, y_1\) and \(x_2, y_2\) is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]. Plugging in the coordinates of points C and E, we have:\[ d_{CE} = \sqrt{(-2 - (-6))^2 + (1 - 3)^2} = \sqrt{(-2 + 6)^2 + (-2)^2} = \sqrt{4^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} \].Thus, the distance from C to E is \(\sqrt{20}\), which approximates to 4.47.
2Step 2: Calculate Distance from D to E
Next, we calculate the distance from point D(3, 0) to point E(-2, 1) using the distance formula. Using the same distance formula:\[ d_{DE} = \sqrt{(-2 - 3)^2 + (1 - 0)^2} = \sqrt{(-5)^2 + 1^2} = \sqrt{25 + 1} = \sqrt{26} \].This gives a calculated distance from D to E of \(\sqrt{26}\), which approximates to 5.10.
3Step 3: Compare the Distances
With the calculated distances, we compare \(\sqrt{20}\) and \(\sqrt{26}\) to determine which is smaller. Since \(\sqrt{20} \approx 4.47\) and \(\sqrt{26} \approx 5.10\), we find that \(\sqrt{20}\) is the smaller distance. Therefore, C is closer to E than D is.
Key Concepts
Coordinate GeometryEuclidean DistanceComparison of Distances
Coordinate Geometry
Coordinate geometry, often referred to as analytical geometry, is the branch of mathematics that uses algebraic equations to describe and understand geometric properties of figures and shapes. It revolutionizes the way we perceive the geometry by providing a bridge between algebra and geometry, allowing us to use coordinates to describe geometric entities such as lines, curves, and polygons.
In coordinate geometry, we use a coordinate system to specify the position of points. The most common system is the Cartesian coordinate system, where we define a plane using two perpendicular axes: the x-axis (horizontal) and the y-axis (vertical). Each point in this plane is given by a pair of numerical coordinates \(x, y\), which describe its unique position.
This approach is particularly powerful because it allows for the precise calculation of distances between points, the slopes of lines, and areas of shapes. Through these calculations, solutions to geometric problems can be developed logically and concisely.
In coordinate geometry, we use a coordinate system to specify the position of points. The most common system is the Cartesian coordinate system, where we define a plane using two perpendicular axes: the x-axis (horizontal) and the y-axis (vertical). Each point in this plane is given by a pair of numerical coordinates \(x, y\), which describe its unique position.
This approach is particularly powerful because it allows for the precise calculation of distances between points, the slopes of lines, and areas of shapes. Through these calculations, solutions to geometric problems can be developed logically and concisely.
Euclidean Distance
The Euclidean distance is a measure of the true straight line distance between two points in Euclidean space. Named after the ancient Greek mathematician Euclid, this concept is foundational in geometry. It represents the "as-the-crow-flies" distance between two points, ignoring any obstacles.
The distance formula derived from the Pythagorean theorem is used to calculate this distance. If we have two points with coordinates \(x_1, y_1\) and \(x_2, y_2\), the Euclidean distance \(d\) between these points can be calculated as:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
This formula allows for finding distances in a coordinate plane with ease and precision, essential for solving problems involving comparisons of distances or even for plotting courses and determining proximities, like the problem of finding which point is closer to another.
The distance formula derived from the Pythagorean theorem is used to calculate this distance. If we have two points with coordinates \(x_1, y_1\) and \(x_2, y_2\), the Euclidean distance \(d\) between these points can be calculated as:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
This formula allows for finding distances in a coordinate plane with ease and precision, essential for solving problems involving comparisons of distances or even for plotting courses and determining proximities, like the problem of finding which point is closer to another.
Comparison of Distances
When dealing with problems that require determining which point is closer to another, it’s essential to compare the calculated distances between pairs of points. After employing the distance formula, as demonstrated in calculating the distances from points C and D to point E, comparing these values helps solve the problem at hand.
Consider the exercise: calculating the distance of point \( C(-6,3) \) and \( D(3,0) \) to the point \( E(-2,1) \). We find:
- Distance \( CE = \sqrt{20} \) approximates to 4.47.
- Distance \( DE = \sqrt{26} \) approximates to 5.10.
By comparing these distances, one can conclude which point lies closer to another by identifying the smaller value. The smaller distance from C to E indicates that point C is closer to E than D is. This comparison is a straightforward application of using computed values to solve spatial problems in coordinate geometry and has various real-world applications, such as optimizing routes in navigation or determining adjacency in networks.
Consider the exercise: calculating the distance of point \( C(-6,3) \) and \( D(3,0) \) to the point \( E(-2,1) \). We find:
- Distance \( CE = \sqrt{20} \) approximates to 4.47.
- Distance \( DE = \sqrt{26} \) approximates to 5.10.
By comparing these distances, one can conclude which point lies closer to another by identifying the smaller value. The smaller distance from C to E indicates that point C is closer to E than D is. This comparison is a straightforward application of using computed values to solve spatial problems in coordinate geometry and has various real-world applications, such as optimizing routes in navigation or determining adjacency in networks.
Other exercises in this chapter
Problem 40
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