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quiz 8.2.3: the distance formulaquestion 1given the coordinates of two points in the plane, you can use either the distance formula or the pythagorean theorem to find the distance between them.a. trueb. falsequestion 2the pythagorean theorem is derived from the distance formula.a. trueb. falsequestion 3the distance between points $(x_1, y_1)$ and $(4, 8)$ is the square root of $(x_1 - 8)^2 + (y_1 - 4)^2$.a. trueb. falsequestion 4which expression gives the distance between the points (4, 6) and (7, -3)?a. $(4 - 7)^2 + (6 + 3)^2$b. $(4 - 7)^2 + (6 - 3)^2$c. $sqrt{(4 - 7)^2 + (6 - 3)^2}$d. $sqrt{(4 - 7)^2 + (6 + 3)^2}$question 5what is the distance between the points (4, 5) and (10, 13) on a coordinate plane?a. 8 unitsb. 12 unitsc. 10 unitsd. 14 unitsquestion 6what is the approximate distance between the points (1, -2) and (-9, 3) on a coordinate grid?a. 8.66 unitsb. 9.43 unitsc. 3.87 unitsd. 11.18 units
Question 1: Verify statement validity
The distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ is derived from the Pythagorean theorem, and both can calculate the distance between two planar points.
Question 2: Verify derivation direction
The distance formula is derived from the Pythagorean theorem, not the reverse.
Question 3: Check distance formula application
The correct distance formula uses differences of matching coordinates: $d=\sqrt{(x_1-4)^2+(y_1-8)^2}$, not swapping $x$ and $y$ terms.
Question 4: Apply distance formula
For points $(4,6)$ and $(7,-3)$, substitute into $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Here, $y_2-y_1=-3-6=-9$, so $(6 - (-3))=6+3$.
Question 5: Calculate exact distance
Substitute $(4,5)$ and $(10,13)$ into the formula:
$d=\sqrt{(10-4)^2+(13-5)^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}$
Question 6: Calculate approximate distance
Substitute $(1,-2)$ and $(-9,3)$ into the formula:
$d=\sqrt{(-9-1)^2+(3-(-2))^2}=\sqrt{(-10)^2+5^2}=\sqrt{100+25}=\sqrt{125}$
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- A. True
- B. False
- B. False
- D. $\sqrt{(4 - 7)^2 + (6 + 3)^2}$
- C. 10 units
- D. 11.18 units