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Completed geometry homework worksheet featuring distance and midpoint formula calculations.

A student's completed worksheet for "Unit 1: Geometry Basics - Homework 3: Distance & Midpoint Formulas," showing handwritten answers to problems involving the distance formula and midpoint coordinates, with a grid for plotting points.

A student's completed worksheet for "Unit 1: Geometry Basics - Homework 3: Distance & Midpoint Formulas," showing handwritten answers to problems involving the distance formula and midpoint coordinates, with a grid for plotting points.

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Show Answer Key & Explanations Step-by-step solution for: SOLUTION: Distance and Midpoint Formulas Worksheet - Studypool

Problem Analysis:


The task involves solving two types of problems:
1. Finding the distance between pairs of points using the distance formula.
2. Finding the midpoint of line segments defined by pairs of points using the midpoint formula.

#### Distance Formula:
The distance \( d \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

#### Midpoint Formula:
The midpoint \( M \) of a line segment with endpoints \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]

Solution:



#### Part 1: Finding the Distance Between Each Pair of Points

1. Points: \((-4, 6)\) and \((3, -7)\)
\[
d = \sqrt{(3 - (-4))^2 + (-7 - 6)^2} = \sqrt{(3 + 4)^2 + (-7 - 6)^2} = \sqrt{7^2 + (-13)^2} = \sqrt{49 + 169} = \sqrt{218}
\]
The provided answer is \(\sqrt{114}\), which is incorrect. The correct answer is:
\[
\boxed{\sqrt{218}}
\]

2. Points: \((-6, -5)\) and \((2, 0)\)
\[
d = \sqrt{(2 - (-6))^2 + (0 - (-5))^2} = \sqrt{(2 + 6)^2 + (0 + 5)^2} = \sqrt{8^2 + 5^2} = \sqrt{64 + 25} = \sqrt{89}
\]
The provided answer is \(\sqrt{84}\), which is incorrect. The correct answer is:
\[
\boxed{\sqrt{89}}
\]

3. Points: \((-1, 4)\) and \((1, -1)\)
\[
d = \sqrt{(1 - (-1))^2 + (-1 - 4)^2} = \sqrt{(1 + 1)^2 + (-1 - 4)^2} = \sqrt{2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}
\]
The provided answer is \(\sqrt{29}\), which is correct:
\[
\boxed{\sqrt{29}}
\]

4. Points: \((0, -8)\) and \((3, 2)\)
\[
d = \sqrt{(3 - 0)^2 + (2 - (-8))^2} = \sqrt{3^2 + (2 + 8)^2} = \sqrt{3^2 + 10^2} = \sqrt{9 + 100} = \sqrt{109}
\]
The provided answer is \(\sqrt{109}\), which is correct:
\[
\boxed{\sqrt{109}}
\]

5. Graphical Distance Calculation:
From the graph, the two points are approximately \((2, 3)\) and \((6, 7)\).
\[
d = \sqrt{(6 - 2)^2 + (7 - 3)^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}
\]
The provided answer is \(\sqrt{12}\), which is incorrect. The correct answer is:
\[
\boxed{4\sqrt{2}}
\]

#### Part 2: Finding the Coordinates of the Midpoint

1. Points: \((6, -4)\) and \((2, 8)\)
\[
M = \left( \frac{6 + 2}{2}, \frac{-4 + 8}{2} \right) = \left( \frac{8}{2}, \frac{4}{2} \right) = (4, 2)
\]
The provided answer is \((2, 2)\), which is incorrect. The correct answer is:
\[
\boxed{(4, 2)}
\]

2. Points: \((9, 1)\) and \((3, 7)\)
\[
M = \left( \frac{9 + 3}{2}, \frac{1 + 7}{2} \right) = \left( \frac{12}{2}, \frac{8}{2} \right) = (6, 4)
\]
The provided answer is \((6, 4)\), which is correct:
\[
\boxed{(6, 4)}
\]

3. Points: \((-3, 7)\) and \((3, -6)\)
\[
M = \left( \frac{-3 + 3}{2}, \frac{7 + (-6)}{2} \right) = \left( \frac{0}{2}, \frac{1}{2} \right) = (0, 0.5)
\]
The provided answer is \((0, -6)\), which is incorrect. The correct answer is:
\[
\boxed{(0, 0.5)}
\]

4. Points: \((7, 4)\) and \((-4, 6)\)
\[
M = \left( \frac{7 + (-4)}{2}, \frac{4 + 6}{2} \right) = \left( \frac{3}{2}, \frac{10}{2} \right) = (1.5, 5)
\]
The provided answer is \((-3, 5)\), which is incorrect. The correct answer is:
\[
\boxed{(1.5, 5)}
\]

Final Answers:


1. \(\boxed{\sqrt{218}}\)
2. \(\boxed{\sqrt{89}}\)
3. \(\boxed{\sqrt{29}}\)
4. \(\boxed{\sqrt{109}}\)
5. \(\boxed{4\sqrt{2}}\)

Midpoints:
1. \(\boxed{(4, 2)}\)
2. \(\boxed{(6, 4)}\)
3. \(\boxed{(0, 0.5)}\)
4. \(\boxed{(1.5, 5)}\)
Parent Tip: Review the logic above to help your child master the concept of geometry distance formula worksheet.
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