Chapter 6: Problem 17
Find the distance between each pair of points. Round to the nearest tenth, if necessary. \(T(6,4), U(2,2)\)
Short Answer
Expert verified
The distance between points \( T \) and \( U \) is approximately 4.5.
Step by step solution
01
Identify the Coordinates
The given points are \( T(6,4) \) and \( U(2,2) \). Identify the coordinates for each point: Point \( T \) has coordinates \((x_1, y_1) = (6, 4)\), and Point \( U \) has coordinates \((x_2, y_2) = (2, 2)\).
02
Apply the Distance Formula
The distance \( d \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Substitute the known coordinates of \( T \) and \( U \) into this formula.
03
Perform Subtraction
Subtract the coordinates: \[ x_2 - x_1 = 2 - 6 = -4 \] \[ y_2 - y_1 = 2 - 4 = -2 \]
04
Square the Differences
Square the differences obtained in the previous step: \[ (-4)^2 = 16 \] \[ (-2)^2 = 4 \]
05
Add the Squares
Add the squared differences: \[ 16 + 4 = 20 \]
06
Calculate the Square Root
Take the square root of the sum from the previous step to find the distance: \[ d = \sqrt{20} \approx 4.4721 \]
07
Round to the Nearest Tenth
Round the distance to the nearest tenth: The distance is approximately \( 4.5 \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Coordinate Geometry
Coordinate geometry is a branch of mathematics that uses a coordinate system to study the position of points, lines, and various geometric shapes on a plane. Here, each point in the plane is described using ordered pairs, often denoted as \((x, y)\). In the context of our problem, the plane is the usual two-dimensional Cartesian plane. Each point's position is determined by its coordinates relative to two perpendicular lines: the x-axis (horizontal) and the y-axis (vertical).
Understanding coordinate geometry is essential for calculating distances, angles, and other properties of geometric figures on the plane. Here are some basics to remember:
Understanding coordinate geometry is essential for calculating distances, angles, and other properties of geometric figures on the plane. Here are some basics to remember:
- The x-coordinate of a point tells you how far along the horizontal axis the point is located.
- The y-coordinate indicates how far along the vertical axis the point is.
- The origin, denoted as \((0, 0)\), is where the axes intersect.
Distance Between Points
To find the distance between two points in a plane, we use the distance formula. This formula is derived from the basic principles of coordinate geometry and helps to calculate the straight-line distance, or Euclidean distance, between two points. The distance formula is:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Where:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Where:
- \( (x_1, y_1) \) are the coordinates of the first point.
- \( (x_2, y_2) \) are the coordinates of the second point.
- Calculate the differences: \( x_2 - x_1 = 2 - 6 = -4 \) and \( y_2 - y_1 = 2 - 4 = -2 \).
- Square these differences: \((-4)^2 = 16\) and \((-2)^2 = 4\).
- Sum up the squares: \(16 + 4 = 20\).
- Finally, take the square root: \(\sqrt{20} \approx 4.4721\), and rounding this gives approximately \(4.5\).
Pythagorean Theorem
The distance formula is closely related to the Pythagorean theorem, a fundamental principle in geometry. The theorem states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. Using triangle terminology:
\[ c^2 = a^2 + b^2 \]
Where:
\[ c^2 = a^2 + b^2 \]
Where:
- \(c\) is the hypotenuse.
- \(a\) and \(b\) are the triangle's other two sides.
- Square each difference, which represents the squares of \(a\) and \(b\).
- Add both squares to find \(c^2\), the square of the hypotenuse.
- The square root of \(c^2\) is \(c\), our desired distance.