Math worksheet with handwritten answers for midpoint and endpoint problems.

Let's go through each problem step by step and verify the solutions. The image contains several geometry problems involving midpoints and segment lengths.

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Problem 10:

> A(4, 4) and B(2, -1), find T if T is the midpoint of AB.

We use the midpoint formula:
$$
T = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
$$

Given:
- A(4, 4)
- B(2, -1)

$$
T_x = \frac{4 + 2}{2} = \frac{6}{2} = 3 \\
T_y = \frac{4 + (-1)}{2} = \frac{3}{2} = 1.5
$$

So, $ T = (3, 1.5) $

But in your answer, you wrote $ T(13, -6) $ — this is incorrect.

✔ Correct Answer: $ T(3, 1.5) $ or $ T\left(3, \frac{3}{2}\right) $

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Problem 11:

> A(-4, -6) and T(7, -2), find R if T is the midpoint of AR.

Let R = (x, y). Since T is the midpoint:

$$
T_x = \frac{-4 + x}{2} = 7 \Rightarrow -4 + x = 14 \Rightarrow x = 18 \\
T_y = \frac{-6 + y}{2} = -2 \Rightarrow -6 + y = -4 \Rightarrow y = 2
$$

So, $ R = (18, 2) $

But you wrote $ R(-7, -4) $ — this is incorrect.

✔ Correct Answer: $ R(18, 2) $

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Problem 12:

> P is the midpoint of AC, and E is the midpoint of BD.
> A(-4, -6), C(1, 4), E(4, -3). Find coordinates of D.

First, find point P (midpoint of AC):

$$
P_x = \frac{-4 + 1}{2} = \frac{-3}{2} = -1.5 \\
P_y = \frac{-6 + 4}{2} = \frac{-2}{2} = -1
\Rightarrow P(-1.5, -1)
$$

Now, E is the midpoint of BD. We know:
- E(4, -3)
- Let D = (x, y)
- B is unknown? Wait — we don’t have B.

Wait — let’s re-read: "P is the midpoint of AC", and "E is the midpoint of BD". But we are given A, C, and E, but not B or D.

Wait — actually, maybe there's a typo or misreading.

Wait — the problem says:
> P is the midpoint of AC and E is the midpoint of BD.
> A(-4,-6), C(1,4), and E(4,-3). Find coordinates of D.

But we don’t know B. So unless B is implied or missing?

Wait — perhaps it's supposed to be that P is the midpoint of AC, and E is the midpoint of BD, and we're told E, but we need more info.

Wait — maybe it's a typo? Or perhaps B is known?

Wait — looking at your answer: you wrote $ D(-3, -7) $

Let’s suppose we are missing B. But without B, we can't find D from E being the midpoint of BD.

Unless... perhaps the diagram implies something else?

Wait — maybe the problem meant:
> P is the midpoint of AC, and E is the midpoint of BD.
> A(-4,-6), C(1,4), and B(-1, 4)? But that's not given.

Wait — maybe it's a different setup.

Alternatively, could it be that P is the midpoint of AC, and E is the midpoint of BD, and we’re told E and A, C, and we need D?

No — still missing B.

Wait — unless “P” is a typo and should be “E”? That doesn’t make sense.

Wait — perhaps it's a quadrilateral, and P and E are midpoints of diagonals?

But no information about B.

Wait — maybe it's a typo, and the point is that E is the midpoint of AD, not BD?

But you wrote D(-3, -7), so let’s test that.

Suppose D = (-3, -7), and E(4, -3) is midpoint of BD. Then we can solve for B.

Let B = (x, y)

Then:
$$
\frac{x + (-3)}{2} = 4 \Rightarrow x - 3 = 8 \Rightarrow x = 11 \\
\frac{y + (-7)}{2} = -3 \Rightarrow y - 7 = -6 \Rightarrow y = 1
\Rightarrow B(11, 1)
$$

But we were not given B, so unless this is consistent with something else...

But we weren’t given any other condition.

Alternatively, maybe the problem is:
> P is the midpoint of AC, and E is the midpoint of BD.
> Given A(-4,-6), C(1,4), and E(4,-3), find D.

But we need B.

Wait — perhaps it's a typo and E is the midpoint of AD, not BD?

Try that:

Let E(4, -3) be midpoint of AD, where A(-4, -6), D(x,y)

Then:
$$
\frac{-4 + x}{2} = 4 \Rightarrow -4 + x = 8 \Rightarrow x = 12 \\
\frac{-6 + y}{2} = -3 \Rightarrow -6 + y = -6 \Rightarrow y = 0
\Rightarrow D(12, 0)
$$

Not matching your answer.

Alternatively, suppose E is midpoint of CD?

C(1,4), D(x,y), E(4,-3)

Then:
$$
\frac{1 + x}{2} = 4 \Rightarrow x = 7 \\
\frac{4 + y}{2} = -3 \Rightarrow y = -10 \Rightarrow D(7, -10)
$$

Still not matching.

Wait — maybe the problem is:
> P is the midpoint of AC, and E is the midpoint of BD.
> A(-4,-6), C(1,4), and E(4,-3), and B is known?

Wait — maybe B is (1, 4)? No, that’s C.

Wait — look at your answer: D(-3, -7)

Let’s suppose that E is the midpoint of BD, and we are to find D, but we don’t know B.

But maybe B is given implicitly?

Wait — perhaps it's a typo and the problem is:

> P is the midpoint of AC, and E is the midpoint of BD.
> A(-4,-6), C(1,4), and B(-1, 4)? Not stated.

Alternatively, perhaps the problem is:
> P is the midpoint of AC, and E is the midpoint of AD.

But again, not matching.

Wait — perhaps the original problem has a diagram where B is known?

Since we don’t have it, and your answer is D(-3, -7), let’s assume that’s what you got.

But based on current info, the problem is incomplete unless B is given.

Wait — another possibility: maybe “P” is a typo, and it's supposed to be that E is the midpoint of AD, and we are to find D?

But then we’d have:

A(-4, -6), E(4, -3), find D.

As above: D(12, 0)

Still not matching.

Wait — maybe it's E is the midpoint of BC, and we need D?

No.

Alternatively, perhaps the problem is:

> P is the midpoint of AC, and E is the midpoint of BD.
> A(-4,-6), C(1,4), and B(-1, 4)? Let's assume B is (-1, 4)

Then E is midpoint of BD, E(4, -3), B(-1,4), find D.

Let D = (x, y)

Then:
$$
\frac{-1 + x}{2} = 4 \Rightarrow -1 + x = 8 \Rightarrow x = 9 \\
\frac{4 + y}{2} = -3 \Rightarrow 4 + y = -6 \Rightarrow y = -10
\Rightarrow D(9, -10)
$$

Still not matching.

Wait — maybe B is (1, -6)? Try:

Let’s suppose your answer D(-3, -7) is correct, and E(4, -3) is midpoint of BD.

Then:
$$
\frac{B_x + (-3)}{2} = 4 \Rightarrow B_x = 11 \\
\frac{B_y + (-7)}{2} = -3 \Rightarrow B_y = 1 \Rightarrow B(11, 1)
$$

Is there any reason to believe B is (11,1)? Not from given data.

So unless there's more context, your answer D(-3, -7) cannot be verified.

But perhaps I misread the problem.

Wait — let me check again.

> P is the midpoint of AC, and E is the midpoint of BD.
> A(-4,-6), C(1,4), and E(4,-3). Find D.

But we don’t know B.

Unless — is it possible that P = E? But P is midpoint of AC, which is:

P = ((-4+1)/2, (-6+4)/2) = (-1.5, -1)

E is (4, -3) → not same.

So not equal.

Thus, this problem is underspecified unless B is known.

Therefore, your answer D(-3, -7) appears incorrect without knowing B.

But since you wrote it, perhaps there was a diagram or typo.

Let’s move on.

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Problem 13:

> PQ = 3x + 14 and QR = 7x - 10. Find PR.

Assuming Q is the midpoint of PR, then:
$$
PQ = QR
$$
So:
$$
3x + 14 = 7x - 10 \\
14 + 10 = 7x - 3x \\
24 = 4x \Rightarrow x = 6
$$

Then:
PQ = 3(6) + 14 = 18 + 14 = 32
QR = 7(6) - 10 = 42 - 10 = 32

So PR = PQ + QR = 32 + 32 = 64

But you wrote PR = 54 — this is incorrect.

✔ Correct Answer: PR = 64

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Problem 14:

> PQ = 2x + 1 and QR = 3x - 44. Find PQ.

Again, Q is midpoint ⇒ PQ = QR

$$
2x + 1 = 3x - 44 \\
1 + 44 = 3x - 2x \\
45 = x
$$

Then PQ = 2(45) + 1 = 90 + 1 = 91

You wrote PQ = 31 — this is incorrect.

✔ Correct Answer: PQ = 91

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Problem 15:

> PQ = 6x + 25 and QR = 14 - 3x. Find PR.

Set PQ = QR (since Q is midpoint):

$$
6x + 25 = 14 - 3x \\
6x + 3x = 14 - 25 \\
9x = -11 \Rightarrow x = -\frac{11}{9}
$$

Then PQ = 6(-11/9) + 25 = -66/9 + 25 = -22/3 + 25 = ( -22 + 75 ) / 3 = 53/3 ≈ 17.67

QR = 14 - 3(-11/9) = 14 + 33/9 = 14 + 11/3 = (42 + 11)/3 = 53/3

So PR = PQ + QR = 53/3 + 53/3 = 106/3 ≈ 35.33

But you wrote PR = 38 — incorrect.

✔ Correct Answer: $ PR = \frac{106}{3} $ or approximately 35.33

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Problem 16:

> PQ = 5x - 31 and QR = 43. Find x.

Q is midpoint ⇒ PQ = QR

So:
$$
5x - 31 = 43 \\
5x = 74 \Rightarrow x = 14.8
$$

But you wrote x = 12 — incorrect.

Check: If x = 12, PQ = 5(12) - 31 = 60 - 31 = 29 ≠ 43 → wrong.

✔ Correct Answer: $ x = \frac{74}{5} = 14.8 $

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✔ Summary of Corrections:

| Problem | Your Answer | Correct Answer |
|--------|-------------|----------------|
| 10 | T(13, -6) | T(3, 1.5) |
| 11 | R(-7, -4) | R(18, 2) |
| 12 | D(-3, -7) | Incomplete (need B) |
| 13 | PR = 54 | PR = 64 |
| 14 | PQ = 31 | PQ = 91 |
| 15 | PR = 38 | PR = 106/3 ≈ 35.33 |
| 16 | x = 12 | x = 14.8 |

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🔍 Final Notes:

- Problems 10–11 involve midpoint formula.
- Problems 13–16 involve midpoint property: if Q is midpoint, then PQ = QR.
- Problem 12 seems missing information (need B or more context).

If you can provide more details about problem 12 (e.g., diagram or full text), I can help resolve it.

Let me know!