Chapter 2: Problem 30
Find the distance \(d\) between the points \(P_{1}\) and \(P_{2}\). $$ P_{1}=(1.2,2.3) ; \quad P_{2}=(-0.3,1.1) $$
Short Answer
Expert verified
The distance \( d \) is 1.92.
Step by step solution
01
Identify the coordinates
The coordinates of the points are given as follows: \( P_{1} = (1.2, 2.3) \) and \( P_{2} = (-0.3, 1.1) \).
02
Write the distance formula
The formula to find the distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
03
Substitute the coordinates into the formula
Substitute \( x_1 = 1.2 \), \( y_1 = 2.3 \), \( x_2 = -0.3 \), and \( y_2 = 1.1 \) into the distance formula: \[ d = \sqrt{((-0.3) - 1.2)^2 + (1.1 - 2.3)^2} \]
04
Simplify inside the square root
First, calculate the differences: \[ -0.3 - 1.2 = -1.5 \] and \[ 1.1 - 2.3 = -1.2 \]
05
Square the differences
Now, square the differences obtained: \[ (-1.5)^2 = 2.25 \] and \[ (-1.2)^2 = 1.44 \]
06
Add the squared differences
Add the squared values: \[ 2.25 + 1.44 = 3.69 \]
07
Compute the square root
Finally, take the square root of the sum: \[ d = \sqrt{3.69} = 1.92 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
coordinate geometry
Coordinate geometry, also known as analytic geometry, allows us to study geometry using a coordinate system. In this case, we use the Cartesian coordinate system to represent points in the plane using pairs of numbers, often referred to as coordinates \( (x, y) \). For example, in the exercise, the points \(P_1 = (1.2, 2.3)\) and \( P_2 = (-0.3, 1.1) \) are given in the Cartesian plane.
This system helps us to visualize geometric problems and provides a way to calculate various properties like distance, slope, and area using algebraic equations and formulas. By understanding the coordinates of the points, we can move on to calculating the distance between these points.
This system helps us to visualize geometric problems and provides a way to calculate various properties like distance, slope, and area using algebraic equations and formulas. By understanding the coordinates of the points, we can move on to calculating the distance between these points.
distance calculation
The distance calculation between two points in a plane is a crucial concept in coordinate geometry. To find the distance between two points, such as \(P_1 \) and \( P_2\), we use the distance formula. This formula is derived from the Pythagorean theorem and is written as:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Here:
Using this formula, we plug in the coordinates from the exercise to get:
\[ d = \sqrt{((-0.3) - 1.2)^2 + (1.1 - 2.3)^2} \]
This formula simplifies not just the calculation but also reduces the possibility of errors.
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Here:
- \( x_1 \) and \( x_2 \) are the x-coordinates of \( P_1 \) and \( P_2 \), respectively.
- \( y_1 \) and \( y_2 \) are the y-coordinates of \( P_1 \) and \( P_2 \).
Using this formula, we plug in the coordinates from the exercise to get:
\[ d = \sqrt{((-0.3) - 1.2)^2 + (1.1 - 2.3)^2} \]
This formula simplifies not just the calculation but also reduces the possibility of errors.
Pythagorean theorem
The distance formula is closely related to the Pythagorean theorem. The Pythagorean theorem applies to right-angled triangles and states:
\[ a^2 + b^2 = c^2 \]
Where:
When finding the distance between two points, we can imagine a right triangle where the horizontal and vertical changes between the points form the legs of this triangle, and the distance between the points is the hypotenuse. So, in the formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
\((x_2 - x_1)\) represents one leg and \((y_2 - y_1)\) represents the other leg of the triangle. Squaring these values and taking their sum follows the Pythagorean theorem, and taking the square root gives us the distance \(d\).
\[ a^2 + b^2 = c^2 \]
Where:
- \( a\) and \( b\) are the lengths of the two legs of a right triangle.
- \( c \) is the length of the hypotenuse, the side opposite the right angle.
When finding the distance between two points, we can imagine a right triangle where the horizontal and vertical changes between the points form the legs of this triangle, and the distance between the points is the hypotenuse. So, in the formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
\((x_2 - x_1)\) represents one leg and \((y_2 - y_1)\) represents the other leg of the triangle. Squaring these values and taking their sum follows the Pythagorean theorem, and taking the square root gives us the distance \(d\).
algebra
Algebra is the branch of mathematics dealing with symbols and the rules for manipulating those symbols. In this context, it helps us to rearrange and solve equations for finding distances.
For example, after substituting the coordinates into the distance formula, we perform algebraic operations to find the distance:
\[ d = \sqrt{((-0.3) - 1.2)^2 + (1.1 - 2.3)^2} \]
We simplify inside the square root by calculating the differences, \((x_2 - x_1) = -1.5\) and \((y_2 - y_1) = -1.2\). Next, we square these values:
\[-1.5^2 = 2.25\]
\[ -1.2^2 = 1.44\]
We add them together:
\[ 2.25 + 1.44 = 3.69 \]
Finally, we take the square root:
\[ d = \sqrt{3.69} = 1.92 \]
By understanding algebraic manipulation, students can confidently solve such problems and verify their results.
For example, after substituting the coordinates into the distance formula, we perform algebraic operations to find the distance:
\[ d = \sqrt{((-0.3) - 1.2)^2 + (1.1 - 2.3)^2} \]
We simplify inside the square root by calculating the differences, \((x_2 - x_1) = -1.5\) and \((y_2 - y_1) = -1.2\). Next, we square these values:
\[-1.5^2 = 2.25\]
\[ -1.2^2 = 1.44\]
We add them together:
\[ 2.25 + 1.44 = 3.69 \]
Finally, we take the square root:
\[ d = \sqrt{3.69} = 1.92 \]
By understanding algebraic manipulation, students can confidently solve such problems and verify their results.
