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Math worksheet for calculating distances between points on a coordinate plane using the distance formula.

A math worksheet titled "Distance on the Coordinate Plane" from Math Monks, featuring nine grids with pairs of points labeled P, Q, C, D, G, H, A, B, E, F, K, L, S, T, I, J. Students are instructed to find the distance between each pair using the distance formula.

A math worksheet titled "Distance on the Coordinate Plane" from Math Monks, featuring nine grids with pairs of points labeled P, Q, C, D, G, H, A, B, E, F, K, L, S, T, I, J. Students are instructed to find the distance between each pair using the distance formula.

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Show Answer Key & Explanations Step-by-step solution for: Distance on the Coordinate Plane Worksheets - Math Monks
Let's solve each distance problem using the distance formula:

$$
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$

We will identify the coordinates of each pair of points from the graph and compute the distances.

---

1. PQ


- Point P: (2, 2)
- Point Q: (5, 5)

$$
PQ = \sqrt{(5 - 2)^2 + (5 - 2)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}
$$

PQ = $3\sqrt{2}$

---

2. CD


- Point C: (-3, 1)
- Point D: (-2, 5)

$$
CD = \sqrt{(-2 - (-3))^2 + (5 - 1)^2} = \sqrt{(1)^2 + (4)^2} = \sqrt{1 + 16} = \sqrt{17}
$$

CD = $\sqrt{17}$

---

3. PQ (Second one)


- Point P: (-2, 2)
- Point Q: (-4, -5)

$$
PQ = \sqrt{(-4 - (-2))^2 + (-5 - 2)^2} = \sqrt{(-2)^2 + (-7)^2} = \sqrt{4 + 49} = \sqrt{53}
$$

PQ = $\sqrt{53}$

---

4. GH


- Point G: (2, -2)
- Point H: (4, -5)

$$
GH = \sqrt{(4 - 2)^2 + (-5 - (-2))^2} = \sqrt{(2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}
$$

GH = $\sqrt{13}$

---

5. AB


- Point A: (-4, -3)
- Point B: (3, -2)

$$
AB = \sqrt{(3 - (-4))^2 + (-2 - (-3))^2} = \sqrt{(7)^2 + (1)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}
$$

AB = $5\sqrt{2}$

---

6. EF


- Point E: (-4, 1)
- Point F: (2, -5)

$$
EF = \sqrt{(2 - (-4))^2 + (-5 - 1)^2} = \sqrt{(6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}
$$

EF = $6\sqrt{2}$

---

7. KL


- Point K: (0, -3)
- Point L: (3, 0)

$$
KL = \sqrt{(3 - 0)^2 + (0 - (-3))^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}
$$

KL = $3\sqrt{2}$

---

8. ST


- Point S: (-5, -2)
- Point T: (0, -4)

$$
ST = \sqrt{(0 - (-5))^2 + (-4 - (-2))^2} = \sqrt{(5)^2 + (-2)^2} = \sqrt{25 + 4} = \sqrt{29}
$$

ST = $\sqrt{29}$

---

9. IJ


- Point I: (0, 4)
- Point J: (1, 3)

$$
IJ = \sqrt{(1 - 0)^2 + (3 - 4)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}
$$

IJ = $\sqrt{2}$

---

Final Answers:



| Segment | Distance |
|--------|----------|
| PQ | $3\sqrt{2}$ |
| CD | $\sqrt{17}$ |
| PQ | $\sqrt{53}$ |
| GH | $\sqrt{13}$ |
| AB | $5\sqrt{2}$ |
| EF | $6\sqrt{2}$ |
| KL | $3\sqrt{2}$ |
| ST | $\sqrt{29}$ |
| IJ | $\sqrt{2}$ |

> ⚠️ Note: The first and third problems both ask for "PQ", but they are different pairs of points. Make sure to label them correctly based on their positions.

Let me know if you'd like this in a printable format or with decimal approximations!
Parent Tip: Review the logic above to help your child master the concept of 6th grade coordinate plane worksheet.
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