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The problem asks to show that the sum of the distances \(|AB| + |BC|\) is equal to the distance \(|AC|\), where the coordinates of points \(A\), \(B\), and \(C\) are given as:

- \(A (-1, 3)\)
- \(B (2, 1)\)
- \(C (5, -1)\)

We can calculate the distances between the points using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Let's compute each distance step by step:

1. **Distance \(AB\):**
   \[
   |AB| = \sqrt{(2 - (-1))^2 + (1 - 3)^2} = \sqrt{(2 + 1)^2 + (1 - 3)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}
   \]

2. **Distance \(BC\):**
   \[
   |BC| = \sqrt{(5 - 2)^2 + (-1 - 1)^2} = \sqrt{(3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}
   \]

3. **Distance \(AC\):**
   \[
   |AC| = \sqrt{(5 - (-1))^2 + (-1 - 3)^2} = \sqrt{(5 + 1)^2 + (-1 - 3)^2} = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52}
   \]
   \[
   \sqrt{52} = 2\sqrt{13}
   \]

Now, sum \( |AB| + |BC| \):
\[
|AB| + |BC| = \sqrt{13} + \sqrt{13} = 2\sqrt{13}
\]

Clearly, \( |AB| + |BC| = |AC| \), which satisfies the equation.

Thus, the distances add up correctly:
\[
|AB| + |BC| = |AC|
\]

Would you like further details or have any questions? Here are some additional related questions:

1. How is the distance formula derived?
2. What is the geometric interpretation of the result \( |AB| + |BC| = |AC| \)?
3. How can this problem be extended to 3D coordinates?
4. Are there other methods to prove the same result?
5. How would this result change if one of the points were shifted?

**Tip:** When calculating distances between points, always double-check the signs of the coordinates to avoid errors.

Show that the sum of the distances |AB| + |BC| equals the distance |AC|, where A(-1, 3), B(2, 1), and C(5, -1).

Learn how to prove that the sum of the distances |AB| + |BC| is equal to |AC| using the distance formula. The points A(-1, 3), B(2, 1), and C(5, -1) are used to illustrate this geometric concept, suitable for Grades 8-10.

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