To find the length of side \( AC \) in the triangle with vertices \( A(-2, -1) \) and \( C(1, -5) \), we can again use the distance formula:

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

For the points \( A(-2, -1) \) and \( C(1, -5) \):

\[
d = \sqrt{(1 - (-2))^2 + (-5 - (-1))^2}
\]

\[
d = \sqrt{(1 + 2)^2 + (-5 + 1)^2}
\]

\[
d = \sqrt{3^2 + (-4)^2}
\]

\[
d = \sqrt{9 + 16}
\]

\[
d = \sqrt{25} = 5
\]

So, the length of \( AC \) is also \( 5 \) units.

Do you have any other questions, or would you like more details?

Here are some related questions to consider:
1. What is the length of side \( BC \)?
2. Can you find the coordinates of the midpoint of side \( AC \)?
3. How do you determine if the triangle is isosceles, equilateral, or scalene?
4. What is the slope of the line passing through points \( A \) and \( C \)?
5. How would you calculate the angle between sides \( AB \) and \( AC \)?

**Tip:** When using the distance formula, make sure to carefully handle negative signs, especially when subtracting coordinates.

Discover how to find the length of side AC in a triangle with vertices at coordinates (-2, -1) and (1, -5). Step-by-step solution using the distance formula provided. Suitable for Grades 9-10.

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