Problem 27
Question
In Exercises 27-38, find the distance between the points. \( (6, -3) \), \( (6, 5) \)
Step-by-Step Solution
Verified Answer
The distance between the points (6, -3) and (6, 5) is 8.
1Step 1: Identifying Given Points
The two points given are (6, -3) and (6, 5). Identify these as (x1, y1) = (6, -3) and (x2, y2) = (6, 5) respectively.
2Step 2: Simplifying the Distance Formula
Because the two points share the same x-coordinate, the distance formula simplifies to the absolute difference of the y-coordinates, i.e., |y2 - y1|. Substituting the given points values into the simplified formula gives: |-3 - 5|.
3Step 3: Calculation
Calculate the absolute value of the expression |-3 - 5| to get: | -8 | which equals 8.
Key Concepts
Coordinate GeometryAbsolute ValueDistance Calculation
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, allows us to determine the position of points in a plane using numerical coordinates. It's like giving every spot on a map a unique set of coordinates, called
"ordered pairs," which look like (x, y). Each pair tells us the point's location on the x (horizontal) and y (vertical) axes. For example, the points given in the exercise are (6, -3) and (6, 5). Here, '6' is the x-coordinate, and it is the same for both points, which means they are vertically aligned on the coordinate plane.
- The x-coordinate tells us how far to move left or right.
- The y-coordinate tells us how far to move up or down.
Absolute Value
Absolute value is a mathematical function that tells us how far a number is from zero on a number line, regardless of direction. It is denoted by two vertical lines, like \(|x|\). The absolute value of a number is never negative.
- For example, both \(|8|\) and \(|-8|\) equal 8.
- Absolute value helps simplify calculations by removing negative signs.
Distance Calculation
Distance calculation in coordinate geometry often uses the distance formula, derived from the Pythagorean theorem. The general formula for calculating the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:\[D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]However, when the points share the same x-coordinate, as in the given exercise, the formula simplifies as there's no horizontal distance:\[D = |y_2 - y_1|\]In our example, the points (6, -3) and (6, 5) only differ in their y-coordinates. Thus, we compute |(-3) - (5)|, resulting in \(8\). Here's why it's simplified:
- Simplification occurs because you're calculating a vertical line segment's length on the plane.
- The distance is a straightforward subtraction of y-coordinates, followed by calculating the absolute value.
Other exercises in this chapter
Problem 27
In Exercises 17-28, find the slope and \(y\)-intercept (if possible) of the equation of the line. Sketch the line. \( x + 5 = 0 \)
View solution Problem 27
In Exercises 23-32, find the \( x \)- and \( y \)-intercepts of the graph of the equation. \( y = |3x-7| \)
View solution Problem 28
In Exercises 27-30, use the given value of \(k\) to complete the table for the inverse variation model \(y = \frac{k}{x^2}\) Plot the points on a rectangular co
View solution Problem 28
In Exercises 17-28, evaluate the indicated function for \(f(x) = x^2 + 1\) and \(g(x) = x - 4\). \((fg)(5)+f(4)\)
View solution