Distance on the Coordinate Plane Worksheets - Math Monks - Free Printable

To solve the problem of finding the distance between pairs of points on the coordinate plane, we will use the distance formula:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

where \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points.

Let's go through each pair of points step by step.

---

1. PQ

- Coordinates of \(P\): \((2, 2)\)
- Coordinates of \(Q\): \((5, 5)\)

Using the distance formula:
\[
d = \sqrt{(5 - 2)^2 + (5 - 2)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}
\]

So, \(PQ = 3\sqrt{2}\).

---

2. CD

- Coordinates of \(C\): \((-4, 0)\)
- Coordinates of \(D\): \((-4, 5)\)

Using the distance formula:
\[
d = \sqrt{(-4 - (-4))^2 + (5 - 0)^2} = \sqrt{0^2 + 5^2} = \sqrt{0 + 25} = \sqrt{25} = 5
\]

So, \(CD = 5\).

---

3. PQ

- Coordinates of \(P\): \((-3, 2)\)
- Coordinates of \(Q\): \((-3, -5)\)

Using the distance formula:
\[
d = \sqrt{(-3 - (-3))^2 + (-5 - 2)^2} = \sqrt{0^2 + (-7)^2} = \sqrt{0 + 49} = \sqrt{49} = 7
\]

So, \(PQ = 7\).

---

4. GH

- Coordinates of \(G\): \((1, -2)\)
- Coordinates of \(H\): \((5, -6)\)

Using the distance formula:
\[
d = \sqrt{(5 - 1)^2 + (-6 - (-2))^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}
\]

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

---

5. AB

- Coordinates of \(A\): \((-3, -4)\)
- Coordinates of \(B\): \((3, -2)\)

Using the distance formula:
\[
d = \sqrt{(3 - (-3))^2 + (-2 - (-4))^2} = \sqrt{(3 + 3)^2 + (-2 + 4)^2} = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}
\]

So, \(AB = 2\sqrt{10}\).

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6. EF

- Coordinates of \(E\): \((-2, -2)\)
- Coordinates of \(F\): \((4, -6)\)

Using the distance formula:
\[
d = \sqrt{(4 - (-2))^2 + (-6 - (-2))^2} = \sqrt{(4 + 2)^2 + (-6 + 2)^2} = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}
\]

So, \(EF = 2\sqrt{13}\).

---

7. KL

- Coordinates of \(K\): \((-1, -2)\)
- Coordinates of \(L\): \((3, 4)\)

Using the distance formula:
\[
d = \sqrt{(3 - (-1))^2 + (4 - (-2))^2} = \sqrt{(3 + 1)^2 + (4 + 2)^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}
\]

So, \(KL = 2\sqrt{13}\).

---

8. ST

- Coordinates of \(S\): \((-5, -2)\)
- Coordinates of \(T\): \((-1, -2)\)

Using the distance formula:
\[
d = \sqrt{(-1 - (-5))^2 + (-2 - (-2))^2} = \sqrt{(-1 + 5)^2 + (-2 + 2)^2} = \sqrt{4^2 + 0^2} = \sqrt{16 + 0} = \sqrt{16} = 4
\]

So, \(ST = 4\).

---

9. IJ

- Coordinates of \(I\): \((-2, 6)\)
- Coordinates of \(J\): \((-2, 4)\)

Using the distance formula:
\[
d = \sqrt{(-2 - (-2))^2 + (4 - 6)^2} = \sqrt{0^2 + (-2)^2} = \sqrt{0 + 4} = \sqrt{4} = 2
\]

So, \(IJ = 2\).

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Final Answers

\[
\boxed{
\begin{aligned}
PQ &= 3\sqrt{2}, & CD &= 5, & PQ &= 7, \\
GH &= 4\sqrt{2}, & AB &= 2\sqrt{10}, & EF &= 2\sqrt{13}, \\
KL &= 2\sqrt{13}, & ST &= 4, & IJ &= 2.
\end{aligned}
}
\]