SOLUTION: Unit 1 Distance and Midpoint Formulas Geometry Basics ... - Free Printable

It looks like you've uploaded an image containing several math problems related to midpoints, coordinates, and distance formulas in coordinate geometry. However, the image is blurry, making it difficult to read all the text clearly.

But based on what I can decipher, here's a breakdown of the problems and their solutions:

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Problem 1: Find the missing endpoint $ C $ given that $ A(3, -4) $ and $ B(2, -7) $, and $ C $ is the midpoint of $ \overline{AB} $.

Wait — this seems incorrect. If $ C $ is the midpoint of $ \overline{AB} $, then we can compute $ C $ using the midpoint formula:

$$
C = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
$$

Given:
- $ A(3, -4) $
- $ B(2, -7) $

So:
$$
x = \frac{3 + 2}{2} = \frac{5}{2} = 2.5
$$
$$
y = \frac{-4 + (-7)}{2} = \frac{-11}{2} = -5.5
$$

So $ C = (2.5, -5.5) $ or $ \left( \frac{5}{2}, -\frac{11}{2} \right) $

But the problem says "Find the missing endpoint $ C $" — so perhaps $ C $ is not the midpoint? Let’s re-read.

Actually, the instruction says:
> "Suppose $ C $ is the midpoint of $ \overline{AB} $", but then gives $ A $ and $ B $, so $ C $ is not missing — it's computable.

Wait — maybe the question is: Given $ A $ and $ C $, find $ B $? Or vice versa?

Let’s look at the next part.

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Problem 2: $ A(4, -6) $, $ C(-1, -4) $, and $ C $ is the midpoint of $ \overline{AB} $. Find $ B $.

This is clearer.

We know:
- $ A(4, -6) $
- $ C(-1, -4) $ is the midpoint
- Let $ B(x, y) $

Using the midpoint formula:

$$
C_x = \frac{A_x + B_x}{2} \Rightarrow -1 = \frac{4 + x}{2}
$$
Multiply both sides by 2:
$$
-2 = 4 + x \Rightarrow x = -6
$$

$$
C_y = \frac{A_y + B_y}{2} \Rightarrow -4 = \frac{-6 + y}{2}
$$
Multiply by 2:
$$
-8 = -6 + y \Rightarrow y = -2
$$

So $ B = (-6, -2) $

✔ Answer: $ \boxed{(-6, -2)} $

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Problem 3: $ A(-1, 2) $, $ B(4, 7) $, and $ C $ is the midpoint of $ \overline{AB} $. Find $ C $.

Use midpoint formula:

$$
C_x = \frac{-1 + 4}{2} = \frac{3}{2} = 1.5
$$
$$
C_y = \frac{2 + 7}{2} = \frac{9}{2} = 4.5
$$

So $ C = \left( \frac{3}{2}, \frac{9}{2} \right) $

✔ Answer: $ \boxed{\left( \frac{3}{2}, \frac{9}{2} \right)} $

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Problem 4: $ A(-1, 2) $, $ C(3, -4) $, and $ C $ is the midpoint of $ \overline{AB} $. Find $ B $.

Let $ B = (x, y) $

Midpoint:
$$
\frac{-1 + x}{2} = 3 \Rightarrow -1 + x = 6 \Rightarrow x = 7
$$
$$
\frac{2 + y}{2} = -4 \Rightarrow 2 + y = -8 \Rightarrow y = -10
$$

So $ B = (7, -10) $

✔ Answer: $ \boxed{(7, -10)} $

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Problem 5: $ P(p) = 2x + 14 $, $ Q(q) = 3x - 10 $, and $ PQ = 28 $. Find $ x $.

Assuming $ P $ and $ Q $ are points on a number line, and $ PQ $ is the distance between them:

$$
|Q - P| = 28
\Rightarrow |(3x - 10) - (2x + 14)| = 28
\Rightarrow |3x - 10 - 2x - 14| = 28
\Rightarrow |x - 24| = 28
$$

Now solve:
$$
x - 24 = 28 \Rightarrow x = 52
$$
or
$$
x - 24 = -28 \Rightarrow x = -4
$$

So two possible values: $ x = 52 $ or $ x = -4 $

But let’s check if both make sense.

If $ x = 52 $:
$ P = 2(52) + 14 = 104 + 14 = 118 $
$ Q = 3(52) - 10 = 156 - 10 = 146 $
Distance = $ |146 - 118| = 28 $ ✔

If $ x = -4 $:
$ P = 2(-4) + 14 = -8 + 14 = 6 $
$ Q = 3(-4) - 10 = -12 - 10 = -22 $
Distance = $ |-22 - 6| = 28 $ ✔

So both are valid.

✔ Answer: $ \boxed{x = 52} $ or $ \boxed{x = -4} $

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Problem 6: $ P(p) = 2x + 1 $, $ Q(q) = 3x - 10 $, $ PQ = 15 $. Find $ x $.

Again, distance:
$$
|(3x - 10) - (2x + 1)| = 15
\Rightarrow |x - 11| = 15
\Rightarrow x - 11 = 15 \Rightarrow x = 26
\quad \text{or} \quad x - 11 = -15 \Rightarrow x = -4
$$

Check:
- $ x = 26 $: $ P = 53 $, $ Q = 78 - 10 = 68 $, $ |68 - 53| = 15 $ ✔
- $ x = -4 $: $ P = -8 + 1 = -7 $, $ Q = -12 - 10 = -22 $, $ |-22 + 7| = 15 $ ✔

✔ Answer: $ \boxed{x = 26} $ or $ \boxed{x = -4} $

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Problem 7: $ P(p) = 4x + 10 $, $ Q(q) = 10 - 3x $, $ PQ = 18 $. Find $ x $.

$$
|Q - P| = 18
\Rightarrow |(10 - 3x) - (4x + 10)| = 18
\Rightarrow |10 - 3x - 4x - 10| = 18
\Rightarrow |-7x| = 18
\Rightarrow 7x = 18 \quad \text{or} \quad -7x = 18
\Rightarrow x = \frac{18}{7} \quad \text{or} \quad x = -\frac{18}{7}
$$

Check:
- $ x = \frac{18}{7} $:
$ P = 4(\frac{18}{7}) + 10 = \frac{72}{7} + \frac{70}{7} = \frac{142}{7} $
$ Q = 10 - 3(\frac{18}{7}) = 10 - \frac{54}{7} = \frac{70 - 54}{7} = \frac{16}{7} $
$ |Q - P| = |\frac{16 - 142}{7}| = \frac{126}{7} = 18 $ ✔

- $ x = -\frac{18}{7} $:
$ P = 4(-\frac{18}{7}) + 10 = -\frac{72}{7} + \frac{70}{7} = -\frac{2}{7} $
$ Q = 10 - 3(-\frac{18}{7}) = 10 + \frac{54}{7} = \frac{70 + 54}{7} = \frac{124}{7} $
$ |Q - P| = |\frac{124}{7} + \frac{2}{7}| = \frac{126}{7} = 18 $ ✔

✔ Answer: $ \boxed{x = \frac{18}{7}} $ or $ \boxed{x = -\frac{18}{7}} $

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Problem 8: $ P(p) = 3x - 10 $, $ Q(q) = 42 $, $ PQ = 20 $. Find $ x $.

$$
|Q - P| = 20 \Rightarrow |42 - (3x - 10)| = 20
\Rightarrow |42 - 3x + 10| = 20
\Rightarrow |52 - 3x| = 20
$$

So:
- $ 52 - 3x = 20 \Rightarrow 3x = 32 \Rightarrow x = \frac{32}{3} $
- $ 52 - 3x = -20 \Rightarrow 3x = 72 \Rightarrow x = 24 $

Check:
- $ x = \frac{32}{3} $: $ P = 3(\frac{32}{3}) - 10 = 32 - 10 = 22 $, $ |42 - 22| = 20 $ ✔
- $ x = 24 $: $ P = 72 - 10 = 62 $, $ |42 - 62| = 20 $ ✔

✔ Answer: $ \boxed{x = \frac{32}{3}} $ or $ \boxed{x = 24} $

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## ✔ Summary of Answers:

| Problem | Answer |
|--------|--------|
| 1 | $ C = \left( \frac{5}{2}, -\frac{11}{2} \right) $ |
| 2 | $ B = (-6, -2) $ |
| 3 | $ C = \left( \frac{3}{2}, \frac{9}{2} \right) $ |
| 4 | $ B = (7, -10) $ |
| 5 | $ x = 52 $ or $ x = -4 $ |
| 6 | $ x = 26 $ or $ x = -4 $ |
| 7 | $ x = \frac{18}{7} $ or $ x = -\frac{18}{7} $ |
| 8 | $ x = \frac{32}{3} $ or $ x = 24 $ |

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If you can clarify which specific problem you'd like explained in detail, or if there's a typo in the original image, I’d be happy to help further!