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Chapter 8: Problem 1
The distance \(d\) from the point (2,-3) to the point (5,1) is _______.
Short Answer
Expert verified
5
Step by step solution
01
- Identify the coordinates
First, identify the coordinates of the two points. The coordinates are (2, -3) and (5, 1).
02
- Write the distance formula
The distance formula for two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
03
- Substitute the coordinates
Substitute the coordinates of the points (2, -3) and (5, 1) into the distance formula: \[ d = \sqrt{(5 - 2)^2 + (1 - ( -3))^2} \]
04
- Calculate the differences
Calculate the differences in the x and y coordinates: \[ (5 - 2) = 3 \] and \[ (1 - (-3)) = 1 + 3 = 4 \]
05
- Square the differences
Square the differences: \[ 3^2 = 9 \] and \[ 4^2 = 16 \]
06
- Sum of squares
Add the squares of the differences: \[ 9 + 16 = 25 \]
07
- Take the square root
Find the square root of the sum: \[ d = \sqrt{25} = 5 \]
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Coordinates
Coordinates are a way to specify the exact location of a point in a plane using a pair of numbers. These numbers are usually written as \((x, y)\).
The first number, called the x-coordinate, specifies the position along the horizontal axis. The second number, or y-coordinate, specifies the position along the vertical axis.
For example, in our exercise, the points \(2, -3 \) and \(5, 1\) are given in coordinate form.
\(2, -3\) means the point is located 2 units to the right along the x-axis and 3 units down along the y-axis.
\(5, 1\) means the point is 5 units to the right along the x-axis and 1 unit up along the y-axis.
Distance Calculation
Distance calculation is a crucial concept in geometry. It tells us how far apart two points are in a plane.
To find the distance between any two points, we use the distance formula. This formula helps by breaking the distance down into simpler mathematical operations.
The formula for the distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
This formula uses the Pythagorean theorem in its calculation, which we'll discuss next.
Pythagorean Theorem
The Pythagorean theorem is fundamental in calculating distances in a plane.
The theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.
It is written as: \[ c^2 = a^2 + b^2 \]
In the distance formula, \((x_2 - x_1)\) and \((y_2 - y_1)\) act as the sides of a right-angled triangle, while \(\text{d}\) is the hypotenuse.
So, using the Pythagorean theorem, we can calculate the distance by finding the hypotenuse between two points.
Squaring Differences
Squaring differences is an essential step in the distance formula.
After finding the differences in the x and y coordinates, we square these values to eliminate any negatives and to follow the rule of the Pythagorean theorem.
In our example, the differences are \(5 - 2 = 3\) and \(1 - (-3) = 4\).
Squaring these differences gives us:
\[ 3^2 = 9 \]
\[ 4^2 = 16 \] This prepares the values for the next step, which is adding them together.
Square Root
Taking the square root is the final step in distance calculation.
After adding the squared differences, we need to find the square root to get the actual distance.
In our example, the sum of squares is \[ 9 + 16 = 25 \]
The square root of 25 is 5, so:
\[ d = \sqrt{25} = 5 \]
The square root function \(\sqrt{}\) reverses the squaring done earlier, giving us a meaningful distance.
This distance of 5 units tells us how far apart the points \(2, -3\) and \(5, 1\) are in the plane.
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