Math worksheet for calculating distance and midpoint of line segments.

Explanation:
We are asked to find two things:
1. The distance between points C and D (using the distance formula or Pythagorean Theorem).
2. The midpoint of segment AB (using the midpoint formula).

Let’s go one problem at a time.

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Distance Formula Reminder:

If you have two points:
C = $(x_1, y_1)$ and D = $(x_2, y_2)$,
then the distance CD is:
$$
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$

This comes from the Pythagorean Theorem — think of the horizontal and vertical differences as legs of a right triangle, and the distance as the hypotenuse.

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Midpoint Formula Reminder:

If A = $(x_1, y_1)$ and B = $(x_2, y_2)$,
then the midpoint M is:
$$
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
$$

Now let’s solve each problem step by step.

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#### 1) C(–3, 4), D(0, –1)

- $x_1 = -3$, $y_1 = 4$
- $x_2 = 0$, $y_2 = -1$

Difference in x: $0 - (-3) = 3$
Difference in y: $-1 - 4 = -5$

Distance = $\sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$
So answer: $\sqrt{34}$

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#### 2) C(–1, 8), D(4, –3)

- $x_1 = -1$, $y_1 = 8$
- $x_2 = 4$, $y_2 = -3$

Δx = $4 - (-1) = 5$
Δy = $-3 - 8 = -11$

Distance = $\sqrt{5^2 + (-11)^2} = \sqrt{25 + 121} = \sqrt{146}$
→ $\sqrt{146}$ (can’t simplify further)

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#### 3) C(16, 7), D(12, –3)

Δx = $12 - 16 = -4$
Δy = $-3 - 7 = -10$

Distance = $\sqrt{(-4)^2 + (-10)^2} = \sqrt{16 + 100} = \sqrt{116}$
Simplify: $116 = 4 \cdot 29$, so $\sqrt{116} = 2\sqrt{29}$

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#### 4) C(4, –1), D(–8, –6)

Δx = $-8 - 4 = -12$
Δy = $-6 - (-1) = -5$

Distance = $\sqrt{(-12)^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13$

Nice! Perfect square.

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#### 5) C(5, 10), D(5, –4)

Same x-coordinate → vertical line.
Distance = |10 – (–4)| = |14| = 14
Or using formula: Δx = 0, Δy = –14 → $\sqrt{0 + 196} = 14$

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#### 6) C(5, –3), D(–7, 2)

Δx = $-7 - 5 = -12$
Δy = $2 - (-3) = 5$

Distance = $\sqrt{(-12)^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$

Again, 13!

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Now midpoints:

#### 7) A(3, –4), B(–3, 0)

Midpoint x: $\frac{3 + (-3)}{2} = \frac{0}{2} = 0$
Midpoint y: $\frac{-4 + 0}{2} = \frac{-4}{2} = -2$
→ Midpoint: $(0, -2)$

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#### 8) A(–2, 1), B(5, 3)

x: $\frac{-2 + 5}{2} = \frac{3}{2} = 1.5$
y: $\frac{1 + 3}{2} = \frac{4}{2} = 2$
→ Midpoint: $(1.5, 2)$ or $\left(\frac{3}{2}, 2\right)$

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#### 9) A(3, –6), B(–9, 2)

x: $\frac{3 + (-9)}{2} = \frac{-6}{2} = -3$
y: $\frac{-6 + 2}{2} = \frac{-4}{2} = -2$
→ Midpoint: $(-3, -2)$

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#### 10) A(13, –6), B(–3, –2)

x: $\frac{13 + (-3)}{2} = \frac{10}{2} = 5$
y: $\frac{-6 + (-2)}{2} = \frac{-8}{2} = -4$
→ Midpoint: $(5, -4)$

Let me double-check all calculations quickly:

1) √34 ✔
2) √146 ✔
3) √116 = 2√29 ✔
4) √169 = 13 ✔
5) 14 ✔
6) √169 = 13 ✔
7) (0, –2) ✔
8) (1.5, 2) ✔
9) (–3, –2) ✔
10) (5, –4) ✔

All correct.

Final Answer:
1) $\sqrt{34}$
2) $\sqrt{146}$
3) $2\sqrt{29}$
4) $13$
5) $14$
6) $13$
7) $(0, -2)$
8) $\left(\frac{3}{2}, 2\right)$
9) $(-3, -2)$
10) $(5, -4)$