Calculate the distance between point (-1, 13) and point (3,-3) in Quadrants II and IV.

To calculate the distance between the points \((-1, 13)\) and \( (3, -3) \), we use the distance formula:

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

In this case:
- \(x_1 = -1\)
- \(y_1 = 13\)
- \(x_2 = 3\)
- \(y_2 = -3\)

Substitute these values into the distance formula:

\[
\text{Distance} = \sqrt{(3 - (-1))^2 + (-3 - 13)^2}
\]

Simplify inside the square root:

\[
\text{Distance} = \sqrt{(3 + 1)^2 + (-3 - 13)^2}
\]

\[
\text{Distance} = \sqrt{4^2 + (-16)^2}
\]

\[
\text{Distance} = \sqrt{16 + 256}
\]

\[
\text{Distance} = \sqrt{272}
\]

Simplify the square root:

\[
\text{Distance} = \sqrt{16 \cdot 17}
\]

\[
\text{Distance} = 4\sqrt{17}
\]

So, the distance between the points \((-1, 13)\) and \( (3, -3) \) is \(4\sqrt{17}\) units.

Would you like further details or have any questions?

Here are some related questions to expand your understanding:

1. What is the distance between the points \((2, 1)\) and \((3, -3)\)?
2. How do you find the midpoint between two points on a coordinate plane?
3. What is the distance between the origin \((0, 0)\) and the point \((3, -3)\)?
4. Can you derive the distance formula from the Pythagorean theorem?
5. What is the distance between \((1, 2)\) and \((2, 1)\)?
6. How do you calculate the distance between two points in 3D space?
7. What is the distance between the points \((-1, -3)\) and \((3, -3)\)?
8. How would the distance formula change if you are working in polar coordinates?

**Tip:** When dealing with large numbers or complex coordinates, break down the calculations into smaller steps to avoid mistakes.

Learn how to calculate the distance between the points (-1, 13) and (3, -3) using the distance formula in coordinate geometry. This problem involves understanding the concept of distance between points in different quadrants on a coordinate plane. Suitable for students in Grades 7-9.

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