Problem 7
Question
find the distance between each pair of points. If necessary, round answers to two decimals places. $$ (-2,-6) \text { and }(3,-4) $$
Step-by-Step Solution
Verified Answer
The distance between the points (-2,-6) and (3,-4) is approximately 5.39.
1Step 1: Identification of coordinates
Identify the coordinates of the two points. In this case, point \(A\) has coordinates \((-2,-6)\) and point \(B\) has coordinates \( (3,-4)\).
2Step 2: Substitute the values into the formula
Substitute the coordinates of the points in the distance formula. This will give \(\sqrt{(3 - (-2))^2 + ((-4) - (-6))^2}\).
3Step 3: Simplify the expression
Simplify the expression inside the square root to get: \(\sqrt{(5)^2 + (2)^2} = \(\sqrt{25 + 4}\).
4Step 4: Calculate the square root
Calculate the square root to find the distance. This will give: \(\sqrt{29}\).
5Step 5: Round off the answer
If necessary, round off to two decimal places. The decimal approximation of \(\sqrt{29}\) to two decimal places is \(5.39\).
Key Concepts
Coordinate GeometryPoint Distance CalculationRounding Decimals
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, is a branch of mathematics that uses coordinate systems to analyze geometric problems. In coordinate geometry, points are represented using ordered pairs \((x, y)\). Each pair describes a location in the two-dimensional plane, defined by a horizontal axis \(x\) and a vertical axis \(y\). This representation allows us to apply algebraic techniques to solve geometric problems.
For example, the points \((-2,-6)\) and \((3,-4)\) are plotted on a graph, where the first value of each pair corresponds to the \((x\)-coordinate, and the second value to the \((y\)-coordinate. By visualizing these points on a coordinate plane, you can observe their relative positions and tackle problems like calculating distances or finding midpoints.
Understanding coordinate geometry is a vital skill as it simplifies and solves complex geometric problems by converting them into algebraic equations.
For example, the points \((-2,-6)\) and \((3,-4)\) are plotted on a graph, where the first value of each pair corresponds to the \((x\)-coordinate, and the second value to the \((y\)-coordinate. By visualizing these points on a coordinate plane, you can observe their relative positions and tackle problems like calculating distances or finding midpoints.
Understanding coordinate geometry is a vital skill as it simplifies and solves complex geometric problems by converting them into algebraic equations.
Point Distance Calculation
Calculating the distance between two points in a coordinate plane is a common problem solved using the distance formula. The formula is derived from the Pythagorean theorem and is given by:
This formula calculates the straight-line distance between two points, also known as the Euclidean distance.
In our example, we find the distance between points \((-2, -6)\) and \(3, -4)\). First, identify the coordinates: \((x_1, y_1) = (-2, -6)\) and \((x_2, y_2) = (3, -4)\). Substitute these values into the formula, \((3 - (-2))^2 + (-4 - (-6))^2\), which simplifies to \[ \sqrt{5^2 + 2^2} = \sqrt{25 + 4} = \sqrt{29} \].
Point distance calculation is practical for measuring the length between two locations in a coordinate space.
- \[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
This formula calculates the straight-line distance between two points, also known as the Euclidean distance.
In our example, we find the distance between points \((-2, -6)\) and \(3, -4)\). First, identify the coordinates: \((x_1, y_1) = (-2, -6)\) and \((x_2, y_2) = (3, -4)\). Substitute these values into the formula, \((3 - (-2))^2 + (-4 - (-6))^2\), which simplifies to \[ \sqrt{5^2 + 2^2} = \sqrt{25 + 4} = \sqrt{29} \].
Point distance calculation is practical for measuring the length between two locations in a coordinate space.
Rounding Decimals
Rounding decimals is crucial when dealing with real-world data where exact precision isn't always feasible or required. In mathematics, rounding helps simplify numbers to make calculations more manageable, and answers easier to understand.
For example, when we calculate the distance between two points and arrive at a value like \(\sqrt{29} \), a precise value isn't always necessary, especially if a simpler approximation suffices. Here, \(\sqrt{29}\approx 5.385\), which can be rounded to two decimal places as 5.39 for simplification.
To round a decimal to two places, inspect the third digit after the decimal point. If it is 5 or more, increase the second digit by one. In our example, 5.385 becomes 5.39. Rounding is a valuable skill for precision where suitably significant digits are required without overwhelming complexity.
For example, when we calculate the distance between two points and arrive at a value like \(\sqrt{29} \), a precise value isn't always necessary, especially if a simpler approximation suffices. Here, \(\sqrt{29}\approx 5.385\), which can be rounded to two decimal places as 5.39 for simplification.
To round a decimal to two places, inspect the third digit after the decimal point. If it is 5 or more, increase the second digit by one. In our example, 5.385 becomes 5.39. Rounding is a valuable skill for precision where suitably significant digits are required without overwhelming complexity.
Other exercises in this chapter
Problem 6
Find the slope of the line passing through each pair of points or state that the slope is undefined. Then indicate whether the line through the points rises fal
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Find the domain of each function. $$g(x)=\frac{3}{x^{2}-2 x-15}$$
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