Coordinate Geometry Notes and Worksheets - Lindsay Bowden - Free Printable

Problem Overview:

The task involves solving problems using the Distance Formula and the Midpoint Formula. Let's break it down step by step.

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Distance Formula

The formula to find the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

Midpoint Formula

The formula to find the midpoint between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:
\[
\left( x, y \right) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]

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Part 1: Distance Formula Problems

#### Problem 1: \((-3, 5)\) and \((-7, -8)\)
\[
d = \sqrt{((-7) - (-3))^2 + ((-8) - 5)^2}
\]
\[
d = \sqrt{(-7 + 3)^2 + (-8 - 5)^2}
\]
\[
d = \sqrt{(-4)^2 + (-13)^2}
\]
\[
d = \sqrt{16 + 169}
\]
\[
d = \sqrt{185}
\]

#### Problem 2: \((-5, 8)\) and \((-4, 1)\)
\[
d = \sqrt{((-4) - (-5))^2 + (1 - 8)^2}
\]
\[
d = \sqrt{(-4 + 5)^2 + (1 - 8)^2}
\]
\[
d = \sqrt{(1)^2 + (-7)^2}
\]
\[
d = \sqrt{1 + 49}
\]
\[
d = \sqrt{50}
\]
\[
d = 5\sqrt{2}
\]

#### Problem 3: \((5, 4)\) and \((3, 8)\)
\[
d = \sqrt{(3 - 5)^2 + (8 - 4)^2}
\]
\[
d = \sqrt{(-2)^2 + (4)^2}
\]
\[
d = \sqrt{4 + 16}
\]
\[
d = \sqrt{20}
\]
\[
d = 2\sqrt{5}
\]

#### Problem 4: \((0, -6)\) and \((-7, 1)\)
\[
d = \sqrt{((-7) - 0)^2 + (1 - (-6))^2}
\]
\[
d = \sqrt{(-7)^2 + (1 + 6)^2}
\]
\[
d = \sqrt{49 + 49}
\]
\[
d = \sqrt{98}
\]
\[
d = 7\sqrt{2}
\]

#### Problem 5: \((2, -3)\) and \((8, -5)\)
\[
d = \sqrt{(8 - 2)^2 + ((-5) - (-3))^2}
\]
\[
d = \sqrt{(8 - 2)^2 + (-5 + 3)^2}
\]
\[
d = \sqrt{(6)^2 + (-2)^2}
\]
\[
d = \sqrt{36 + 4}
\]
\[
d = \sqrt{40}
\]
\[
d = 2\sqrt{10}
\]

#### Problem 6: \((-4, -2)\) and \((7, -3)\)
\[
d = \sqrt{(7 - (-4))^2 + ((-3) - (-2))^2}
\]
\[
d = \sqrt{(7 + 4)^2 + (-3 + 2)^2}
\]
\[
d = \sqrt{(11)^2 + (-1)^2}
\]
\[
d = \sqrt{121 + 1}
\]
\[
d = \sqrt{122}
\]

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Part 2: Midpoint Formula Problems

#### Problem 1: \((8, 0)\) and \((-9, 1)\)
\[
\left( x, y \right) = \left( \frac{8 + (-9)}{2}, \frac{0 + 1}{2} \right)
\]
\[
\left( x, y \right) = \left( \frac{-1}{2}, \frac{1}{2} \right)
\]
\[
\left( x, y \right) = \left( -0.5, 0.5 \right)
\]

#### Problem 2: \((-9, -10)\) and \((4, -9)\)
\[
\left( x, y \right) = \left( \frac{-9 + 4}{2}, \frac{-10 + (-9)}{2} \right)
\]
\[
\left( x, y \right) = \left( \frac{-5}{2}, \frac{-19}{2} \right)
\]
\[
\left( x, y \right) = \left( -2.5, -9.5 \right)
\]

#### Problem 3: \((4, -3)\) and \((2, 9)\)
\[
\left( x, y \right) = \left( \frac{4 + 2}{2}, \frac{-3 + 9}{2} \right)
\]
\[
\left( x, y \right) = \left( \frac{6}{2}, \frac{6}{2} \right)
\]
\[
\left( x, y \right) = \left( 3, 3 \right)
\]

#### Problem 4: \((8, 7)\) and \((-8, 3)\)
\[
\left( x, y \right) = \left( \frac{8 + (-8)}{2}, \frac{7 + 3}{2} \right)
\]
\[
\left( x, y \right) = \left( \frac{0}{2}, \frac{10}{2} \right)
\]
\[
\left( x, y \right) = \left( 0, 5 \right)
\]

#### Problem 5: \((7, -8)\) and \((-2, 8)\)
\[
\left( x, y \right) = \left( \frac{7 + (-2)}{2}, \frac{-8 + 8}{2} \right)
\]
\[
\left( x, y \right) = \left( \frac{5}{2}, \frac{0}{2} \right)
\]
\[
\left( x, y \right) = \left( 2.5, 0 \right)
\]

#### Problem 6: \((-4, 7)\) and \((3, -1)\)
\[
\left( x, y \right) = \left( \frac{-4 + 3}{2}, \frac{7 + (-1)}{2} \right)
\]
\[
\left( x, y \right) = \left( \frac{-1}{2}, \frac{6}{2} \right)
\]
\[
\left( x, y \right) = \left( -0.5, 3 \right)
\]

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Final Answers:

#### Distance Formula Answers:
1. \(\sqrt{185}\)
2. \(5\sqrt{2}\)
3. \(2\sqrt{5}\)
4. \(7\sqrt{2}\)
5. \(2\sqrt{10}\)
6. \(\sqrt{122}\)

#### Midpoint Formula Answers:
1. \((-0.5, 0.5)\)
2. \((-2.5, -9.5)\)
3. \((3, 3)\)
4. \((0, 5)\)
5. \((2.5, 0)\)
6. \((-0.5, 3)\)

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Boxed Final Answer:

\[
\boxed{
\begin{array}{ll}
\text{Distance:} & 1. \sqrt{185}, \quad 2. 5\sqrt{2}, \quad 3. 2\sqrt{5}, \quad 4. 7\sqrt{2}, \quad 5. 2\sqrt{10}, \quad 6. \sqrt{122} \\
\text{Midpoint:} & 1. (-0.5, 0.5), \quad 2. (-2.5, -9.5), \quad 3. (3, 3), \quad 4. (0, 5), \quad 5. (2.5, 0), \quad 6. (-0.5, 3)
\end{array}
}
\]