Find Distance Using Pythagorean Theorem - 8th Grade Math

1. Identify the value of $(y_2-y_1)^2$ when $y_1=10$ and $y_2=4$.

   $36$. $(4-10)^2=(-6)^2=36$

2. Find and correct the error: $d=\sqrt{(x_2-x_1)+(y_2-y_1)}$ for distance between two points.

   $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Missing squares; must square each difference before adding.

3. Find the distance between $(5,9)$ and $(1,0)$ in simplest radical form.

   $\sqrt{97}$. $\sqrt{4^2+9^2}=\sqrt{16+81}=\sqrt{97}$

4. Find the distance between $(-1,-2)$ and $(2,2)$ in simplest radical form.

   $5$. $\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

5. Identify the vertical distance (rise) formula between points $ ($x_1$, $y_1$) $ and $ ($x_2$, $y_2$) $.

   $|y_2-y_1|$. Absolute value ensures positive vertical distance.

6. Identify the error: $d=\sqrt{(x_2-x_1)+(y_2-y_1)}$; state the corrected formula.

   $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Missing squares; must square each difference.

7. Find the distance between $(3,-1)$ and $(-1,2)$.

   $5$. $\sqrt{(-1-3)^2+(2-(-1))^2}=\sqrt{16+9}=\sqrt{25}=5$

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

   $5$. $\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

9. Find the distance between $(-3,-4)$ and $(1,8)$ in simplest radical form.

   $\sqrt{160}$. $\sqrt{4^2+12^2}=\sqrt{16+144}=\sqrt{160}$

10. Identify the error: $d=\sqrt{(x_2-x_1)+(y_2-y_1)}$ is written for distance. What is the correct formula?

    $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Missing squares on the differences inside the square root.

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

    $5$. $\sqrt{(2-6)^2+(2-(-1))^2}=\sqrt{16+9}=5$

12. Which statement is always true about distance between two points: $d$ or $-d$ can be a distance?

    $d\ge 0$. Distance is always non-negative by definition.

13. Identify the value of $(x_2-x_1)^2$ when $x_1=-3$ and $x_2=5$.

    $64$. $(5-(-3))^2=8^2=64$

14. What is the distance between $(1,2)$ and $(4,6)$?

    $5$. $\sqrt{(4-1)^2+(6-2)^2}=\sqrt{9+16}=5$

15. Find the distance between $(-1,-1)$ and $(2,3)$.

    $5$. $\sqrt{(2-(-1))^2+(3-(-1))^2}=\sqrt{9+16}=\sqrt{25}=5$

16. Find the distance between $(1,2)$ and $(5,5)$.

    $5$. $\sqrt{(5-1)^2+(5-2)^2}=\sqrt{16+9}=\sqrt{25}=5$

17. What is the distance between $(-2,3)$ and $(4,-5)$?

    $10$. $\sqrt{(4-(-2))^2+(-5-3)^2}=\sqrt{36+64}=10$

18. What is the vertical distance between $(x_1,y_1)$ and $(x_2,y_2)$ on a coordinate plane?

    $|y_2-y_1|$. The absolute value ensures distance is always positive.

19. Identify the horizontal distance (run) formula between points $(x_1,y_1)$ and $(x_2,y_2)$.

    $|x_2-x_1|$. Absolute value ensures positive horizontal distance.

20. Identify the legs of the right triangle used to find distance between $(x_1,y_1)$ and $(x_2,y_2)$.

    Legs: $|x_2-x_1|$ and $|y_2-y_1|$. Horizontal and vertical distances form perpendicular sides.

21. Find the distance between $(2,1)$ and $(2,6)$.

    $5$. Vertical line, so distance is $|6-1|=5$.

22. Find the distance between $(-6,2)$ and $(3,-10)$ in simplest radical form.

    $15$. $\sqrt{9^2+12^2}=\sqrt{81+144}=\sqrt{225}=15$

23. What is the distance between $(-5,1)$ and $(-2,5)$?

    $5$. $\sqrt{(-2-(-5))^2+(5-1)^2}=\sqrt{9+16}=5$

24. Find the distance between $(0,0)$ and $(6,8)$ using the Pythagorean Theorem.

    $10$. $\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$

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

    $5$. $\sqrt{(2-(-1))^2+(2-(-2))^2}=\sqrt{9+16}=5$

26. What is the distance between $(3,3)$ and $(7,0)$?

    $5$. $\sqrt{(7-3)^2+(0-3)^2}=\sqrt{16+9}=5$

27. Find the distance between $(-1,3)$ and $(4,3)$.

    $5$. Horizontal line, so distance is $|4-(-1)|=5$.

28. What is the distance between $(1,5)$ and $(4,1)$?

    $5$. $\sqrt{(4-1)^2+(1-5)^2}=\sqrt{9+16}=5$

29. Find the distance between $(2,3)$ and $(2,11)$ in simplest form.

    $8$. Vertical line, so distance is $|11-3|=8$.

30. Find the distance between $(-3,4)$ and $(1,1)$.

    $5$. $\sqrt{(1-(-3))^2+(1-4)^2}=\sqrt{16+9}=\sqrt{25}=5$

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

    $5$. $\sqrt{(0-(-4))^2+(0-(-3))^2}=\sqrt{16+9}=5$

32. Find the distance between $(-2,-1)$ and $(1,3)$.

    $5$. $\sqrt{(1-(-2))^2+(3-(-1))^2}=\sqrt{9+16}=\sqrt{25}=5$

33. Find the distance between $(2,-3)$ and $(8,1)$ in simplest radical form.

    $\sqrt{52}$. $\sqrt{6^2+4^2}=\sqrt{36+16}=\sqrt{52}$

34. Which value equals the squared distance between $(1,2)$ and $(5,5)$?

    $25$. $(5-1)^2+(5-2)^2=16+9=25$ before taking square root.

35. What is the distance between $(-3,5)$ and $(4,5)$?

    $7$. Horizontal line, so distance is $|4-(-3)|=7$.

36. Find the distance between $(-4,-3)$ and $(0,0)$.

    $5$. $\sqrt{(-4)^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5$

37. Find the distance between $(-2,5)$ and $(4,1)$ in simplest radical form.

    $\sqrt{52}$. $\sqrt{6^2+(-4)^2}=\sqrt{36+16}=\sqrt{52}$

38. Find the distance between $(7,-2)$ and $(1,2)$.

    $\sqrt{52}=2\sqrt{13}$. $\sqrt{(1-7)^2+(2-(-2))^2}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}$

39. Find the distance between $(-2,5)$ and $(4,1)$.

    $\sqrt{52}=2\sqrt{13}$. $\sqrt{(4-(-2))^2+(1-5)^2}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}$

40. What is the distance between $(1,-1)$ and $(5,2)$?

    $5$. $\sqrt{(5-1)^2+(2-(-1))^2}=\sqrt{16+9}=5$

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

    $5$. $\sqrt{(1-(-2))^2+(3-(-1))^2}=\sqrt{9+16}=5$

42. Find the distance between $(-4,7)$ and $(5,7)$ in simplest form.

    $9$. Horizontal line, so distance is $|5-(-4)|=9$.

43. What is the vertical distance between $(x_1,y_1)$ and $(x_1,y_2)$?

    $|y_2-y_1|$. Points share same x-coordinate, so distance is vertical difference.

44. Find the distance between $(0,0)$ and $(6,8)$.

    $10$. $\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$

45. What is the horizontal distance between $(x_1,y_1)$ and $(x_2,y_1)$?

    $|x_2-x_1|$. Points share same y-coordinate, so distance is horizontal difference.

46. Find the distance between $(1,2)$ and $(4,6)$.

    $5$. $\sqrt{(4-1)^2+(6-2)^2}=\sqrt{9+16}=\sqrt{25}=5$

47. What is the horizontal distance between $(x_1,y_1)$ and $(x_2,y_2)$ on a coordinate plane?

    $|x_2-x_1|$. The absolute value ensures distance is always positive.

48. State the Pythagorean Theorem used to connect legs $a,b$ and hypotenuse $c$.

    $a^2+b^2=c^2$. Relates the squares of a right triangle's sides.

49. Which expression correctly represents the distance between $(a,b)$ and $(c,d)$?

    $\sqrt{(c-a)^2+(d-b)^2}$. Standard distance formula with variables $(a,b)$ and $(c,d)$.

50. Which expression correctly gives the distance between $(x_1,y_1)$ and $(x_2,y_2)$ using Pythagorean Theorem?

    $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Same as distance formula without the $d=$.

51. State the distance formula between points $(x_1,y_1)$ and $(x_2,y_2)$ on a coordinate plane.

    $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Derived from Pythagorean theorem applied to coordinate differences.

52. Find the distance between $(3,-1)$ and $(-1,2)$ in simplest radical form.

    $5$. $\sqrt{(-4)^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$

53. State the distance formula between $(x_1,y_1)$ and $(x_2,y_2)$ in the coordinate plane.

    $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Derived from the Pythagorean theorem with legs as horizontal and vertical distances.

54. What is the distance between $ (0,0) $ and $ (6,8) $?

    $10$. $\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$

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

    $7$. Vertical line, so distance is $|8-1|=7$.

56. Identify the Pythagorean relationship used to find distance with legs $\Delta x$ and $\Delta y$.

    $d^2=(\Delta x)^2+(\Delta y)^2$. Pythagorean theorem: hypotenuse squared equals sum of legs squared.

57. Find the distance between $(1,2)$ and $(7,5)$ in simplest radical form.

    $\sqrt{45}$. $\sqrt{6^2+3^2}=\sqrt{36+9}=\sqrt{45}$

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