Pythagorean Theorem: Distance - Pre Algebra

1. What is the distance between $(-3, 4)$ and $(-3, -2)$?

   6. The points lie on a vertical line, so subtract the $y$-coordinates: $4 - (-2) = 6$.

2. What is the distance between $(-2, 1)$ and $(3, 13)$?

   13. The legs are 5 and 12, and $\sqrt{25 + 144} = \sqrt{169}$.

3. What is the distance formula for points $(x_1, y_1)$ and $(x_2, y_2)$?

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

4. What is the distance between $(1, 2)$ and $(7, 10)$?

   10. The legs are 6 and 8, and $\sqrt{36 + 64} = \sqrt{100}$.

5. How does the distance formula come from the Pythagorean Theorem?

   The horizontal and vertical changes are the legs of a right triangle, and the distance between the points is its hypotenuse.

6. What is the distance between $(-1, -1)$ and $(2, 3)$?

   5. The legs are 3 and 4, and $\sqrt{9 + 16} = \sqrt{25}$.

7. What is the distance between $(5, 7)$ and $(2, 5)$?

   $\sqrt{13}$. The legs are 3 and 2, and $\sqrt{9 + 4} = \sqrt{13}$.

8. What is the distance between $(1, 3)$ and $(5, 1)$?

   $\sqrt{20}$, which simplifies to $2\sqrt{5}$. The legs are 4 and 2, and $16 + 4 = 20$.

9. What is the distance between $(2, 1)$ and $(8, 1)$?

   6. The points lie on a horizontal line, so subtract the $x$-coordinates: $8 - 2 = 6$.

10. What is the distance between $(0, 0)$ and $(1, 1)$?

    $\sqrt{2}$, since $\sqrt{1^2 + 1^2} = \sqrt{2}$.

11. What is the distance between $(0, 0)$ and $(3, 4)$?

    5, since $\sqrt{3^2 + 4^2} = \sqrt{25}$.

12. Why does it not matter which point is called $(x_1, y_1)$ in the distance formula?

    The differences are squared, so a negative difference gives the same square as a positive one.

Study these flashcards