Practice worksheet for finding distances between points on a coordinate plane using the Pythagorean Theorem.

To solve the problem of finding the distance between each pair of points using the Pythagorean Theorem, we will follow these steps:

Step 1: Understand the Pythagorean Theorem

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

Step 2: Identify the coordinates of the points for each pair

We will determine the coordinates of the points from the given graphs and then apply the formula.

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Problem (1): Distance \( AB \)

- Coordinates of \( A \): \( (-2, -2) \)
- Coordinates of \( B \): \( (5, 3) \)

Using the distance formula:
\[
AB = \sqrt{(5 - (-2))^2 + (3 - (-2))^2} = \sqrt{(5 + 2)^2 + (3 + 2)^2} = \sqrt{7^2 + 5^2} = \sqrt{49 + 25} = \sqrt{74}
\]

So, \( AB = \sqrt{74} \).

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Problem (2): Distance \( QR \)

- Coordinates of \( Q \): \( (3, 3) \)
- Coordinates of \( R \): \( (6, 5) \)

Using the distance formula:
\[
QR = \sqrt{(6 - 3)^2 + (5 - 3)^2} = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}
\]

So, \( QR = \sqrt{13} \).

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Problem (3): Distance \( XY \)

- Coordinates of \( X \): \( (2, -3) \)
- Coordinates of \( Y \): \( (2, 2) \)

Using the distance formula:
\[
XY = \sqrt{(2 - 2)^2 + (2 - (-3))^2} = \sqrt{0^2 + (2 + 3)^2} = \sqrt{0 + 5^2} = \sqrt{25} = 5
\]

So, \( XY = 5 \).

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Problem (4): Distance \( PQ \)

- Coordinates of \( P \): \( (-2, -3) \)
- Coordinates of \( Q \): \( (6, 2) \)

Using the distance formula:
\[
PQ = \sqrt{(6 - (-2))^2 + (2 - (-3))^2} = \sqrt{(6 + 2)^2 + (2 + 3)^2} = \sqrt{8^2 + 5^2} = \sqrt{64 + 25} = \sqrt{89}
\]

So, \( PQ = \sqrt{89} \).

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Problem (5): Distance \( ST \)

- Coordinates of \( S \): \( (2, -2) \)
- Coordinates of \( T \): \( (5, -4) \)

Using the distance formula:
\[
ST = \sqrt{(5 - 2)^2 + (-4 - (-2))^2} = \sqrt{3^2 + (-4 + 2)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}
\]

So, \( ST = \sqrt{13} \).

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Problem (6): Distance \( BC \)

- Coordinates of \( B \): \( (-2, -2) \)
- Coordinates of \( C \): \( (2, 2) \)

Using the distance formula:
\[
BC = \sqrt{(2 - (-2))^2 + (2 - (-2))^2} = \sqrt{(2 + 2)^2 + (2 + 2)^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}
\]

So, \( BC = 4\sqrt{2} \).

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Final Answers

\[
\boxed{\sqrt{74}, \sqrt{13}, 5, \sqrt{89}, \sqrt{13}, 4\sqrt{2}}
\]