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8th Grade Math Flashcards: Find Distance Using Pythagorean Theorem

Study Find Distance Using Pythagorean Theorem in 8th Grade Math with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Find Distance Using Pythagorean Theorem, giving you a quick way to review the definitions, rules, and examples that matter most for 8th Grade Math.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

8th Grade Math Flashcards: Find Distance Using Pythagorean Theorem

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QUESTION

What is the distance between (−4,−3)(-4,-3)(−4,−3) and (0,0)(0,0)(0,0)?

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ANSWER

555. (0−(−4))2+(0−(−3))2=16+9=5\sqrt{(0-(-4))^2+(0-(-3))^2}=\sqrt{16+9}=5(0−(−4))2+(0−(−3))2​=16+9​=5

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Flashcard 1: What is the distance between (−4,−3)(-4,-3)(−4,−3) and (0,0)(0,0)(0,0)?

Answer: 555. (0−(−4))2+(0−(−3))2=16+9=5\sqrt{(0-(-4))^2+(0-(-3))^2}=\sqrt{16+9}=5(0−(−4))2+(0−(−3))2​=16+9​=5

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

Answer: 45\sqrt{45}45​. 62+32=36+9=45\sqrt{6^2+3^2}=\sqrt{36+9}=\sqrt{45}62+32​=36+9​=45​

Flashcard 3: Find the distance between (−3,−4)(-3,-4)(−3,−4) and (1,8)(1,8)(1,8) in simplest radical form.

Answer: 160\sqrt{160}160​. 42+122=16+144=160\sqrt{4^2+12^2}=\sqrt{16+144}=\sqrt{160}42+122​=16+144​=160​

Flashcard 4: Identify the Pythagorean relationship used to find distance with legs Δx\Delta xΔx and Δy\Delta yΔy.

Answer: d2=(Δx)2+(Δy)2d^2=(\Delta x)^2+(\Delta y)^2d2=(Δx)2+(Δy)2. Pythagorean theorem: hypotenuse squared equals sum of legs squared.

Flashcard 5: Find the distance between (0,0)(0,0)(0,0) and (6,8)(6,8)(6,8) using the Pythagorean Theorem.

Answer: 101010. 62+82=36+64=100=10\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=1062+82​=36+64​=100​=10

Flashcard 6: Find the distance between (2,3)(2,3)(2,3) and (2,11)(2,11)(2,11) in simplest form.

Answer: 888. Vertical line, so distance is ∣11−3∣=8|11-3|=8∣11−3∣=8.

Flashcard 7: Find the distance between (−4,7)(-4,7)(−4,7) and (5,7)(5,7)(5,7) in simplest form.

Answer: 999. Horizontal line, so distance is ∣5−(−4)∣=9|5-(-4)|=9∣5−(−4)∣=9.

Flashcard 8: Find the distance between (−1,−2)(-1,-2)(−1,−2) and (2,2)(2,2)(2,2) in simplest radical form.

Answer: 555. 32+42=9+16=25=5\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=532+42​=9+16​=25​=5

Flashcard 9: Find the distance between (3,−1)(3,-1)(3,−1) and (−1,2)(-1,2)(−1,2) in simplest radical form.

Answer: 555. (−4)2+32=16+9=25=5\sqrt{(-4)^2+3^2}=\sqrt{16+9}=\sqrt{25}=5(−4)2+32​=16+9​=25​=5

Flashcard 10: Find the distance between (−2,5)(-2,5)(−2,5) and (4,1)(4,1)(4,1) in simplest radical form.

Answer: 52\sqrt{52}52​. 62+(−4)2=36+16=52\sqrt{6^2+(-4)^2}=\sqrt{36+16}=\sqrt{52}62+(−4)2​=36+16​=52​

Flashcard 11: Find the distance between (−6,2)(-6,2)(−6,2) and (3,−10)(3,-10)(3,−10) in simplest radical form.

Answer: 151515. 92+122=81+144=225=15\sqrt{9^2+12^2}=\sqrt{81+144}=\sqrt{225}=1592+122​=81+144​=225​=15

Flashcard 12: Find the distance between (2,−3)(2,-3)(2,−3) and (8,1)(8,1)(8,1) in simplest radical form.

Answer: 52\sqrt{52}52​. 62+42=36+16=52\sqrt{6^2+4^2}=\sqrt{36+16}=\sqrt{52}62+42​=36+16​=52​

Flashcard 13: Identify the value of (y2−y1)2(y_2-y_1)^2(y2​−y1​)2 when y1=10y_1=10y1​=10 and y2=4y_2=4y2​=4.

Answer: 363636. (4−10)2=(−6)2=36(4-10)^2=(-6)^2=36(4−10)2=(−6)2=36

Flashcard 14: Identify the value of (x2−x1)2(x_2-x_1)^2(x2​−x1​)2 when x1=−3x_1=-3x1​=−3 and x2=5x_2=5x2​=5.

Answer: 646464. (5−(−3))2=82=64(5-(-3))^2=8^2=64(5−(−3))2=82=64

Flashcard 15: Which expression correctly represents the distance between (a,b)(a,b)(a,b) and (c,d)(c,d)(c,d)?

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

Flashcard 16: Find the distance between (5,9)(5,9)(5,9) and (1,0)(1,0)(1,0) in simplest radical form.

Answer: 97\sqrt{97}97​. 42+92=16+81=97\sqrt{4^2+9^2}=\sqrt{16+81}=\sqrt{97}42+92​=16+81​=97​

Flashcard 17: State the distance formula between (x1,y1)(x_1,y_1)(x1​,y1​) and (x2,y2)(x_2,y_2)(x2​,y2​) in the coordinate plane.

Answer: d=(x2−x1)2+(y2−y1)2d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}d=(x2​−x1​)2+(y2​−y1​)2​. Derived from the Pythagorean theorem with legs as horizontal and vertical distances.

Flashcard 18: Which statement is always true about distance between two points: ddd or −d-d−d can be a distance?

Answer: d≥0d\ge 0d≥0. Distance is always non-negative by definition.

Flashcard 19: Find and correct the error: d=(x2−x1)+(y2−y1)d=\sqrt{(x_2-x_1)+(y_2-y_1)}d=(x2​−x1​)+(y2​−y1​)​ for distance between two points.

Answer: d=(x2−x1)2+(y2−y1)2d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}d=(x2​−x1​)2+(y2​−y1​)2​. Missing squares; must square each difference before adding.

Flashcard 20: What is the horizontal distance between (x1,y1)(x_1,y_1)(x1​,y1​) and (x2,y2)(x_2,y_2)(x2​,y2​) on a coordinate plane?

Answer: ∣x2−x1∣|x_2-x_1|∣x2​−x1​∣. The absolute value ensures distance is always positive.

Flashcard 21: What is the vertical distance between (x1,y1)(x_1,y_1)(x1​,y1​) and (x2,y2)(x_2,y_2)(x2​,y2​) on a coordinate plane?

Answer: ∣y2−y1∣|y_2-y_1|∣y2​−y1​∣. The absolute value ensures distance is always positive.

Flashcard 22: State the distance formula between points (x1,y1)(x_1,y_1)(x1​,y1​) and (x2,y2)(x_2,y_2)(x2​,y2​) on a coordinate plane.

Answer: d=(x2−x1)2+(y2−y1)2d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}d=(x2​−x1​)2+(y2​−y1​)2​. Derived from Pythagorean theorem applied to coordinate differences.

Flashcard 23: What is the horizontal distance between (x1,y1)(x_1,y_1)(x1​,y1​) and (x2,y1)(x_2,y_1)(x2​,y1​)?

Answer: ∣x2−x1∣|x_2-x_1|∣x2​−x1​∣. Points share same y-coordinate, so distance is horizontal difference.

Flashcard 24: What is the vertical distance between (x1,y1)(x_1,y_1)(x1​,y1​) and (x1,y2)(x_1,y_2)(x1​,y2​)?

Answer: ∣y2−y1∣|y_2-y_1|∣y2​−y1​∣. Points share same x-coordinate, so distance is vertical difference.

Flashcard 25: Identify the legs of the right triangle used to find distance between (x1,y1)(x_1,y_1)(x1​,y1​) and (x2,y2)(x_2,y_2)(x2​,y2​).

Answer: Legs: ∣x2−x1∣|x_2-x_1|∣x2​−x1​∣ and ∣y2−y1∣|y_2-y_1|∣y2​−y1​∣. Horizontal and vertical distances form perpendicular sides.

Flashcard 26: Find the distance between (3,−1)(3,-1)(3,−1) and (−1,2)(-1,2)(−1,2).

Answer: 555. (−1−3)2+(2−(−1))2=16+9=25=5\sqrt{(-1-3)^2+(2-(-1))^2}=\sqrt{16+9}=\sqrt{25}=5(−1−3)2+(2−(−1))2​=16+9​=25​=5

Flashcard 27: Find the distance between (−1,−1)(-1,-1)(−1,−1) and (2,3)(2,3)(2,3).

Answer: 555. (2−(−1))2+(3−(−1))2=9+16=25=5\sqrt{(2-(-1))^2+(3-(-1))^2}=\sqrt{9+16}=\sqrt{25}=5(2−(−1))2+(3−(−1))2​=9+16​=25​=5

Flashcard 28: Find the distance between (1,2)(1,2)(1,2) and (5,5)(5,5)(5,5).

Answer: 555. (5−1)2+(5−2)2=16+9=25=5\sqrt{(5-1)^2+(5-2)^2}=\sqrt{16+9}=\sqrt{25}=5(5−1)2+(5−2)2​=16+9​=25​=5

Flashcard 29: Which value equals the squared distance between (1,2)(1,2)(1,2) and (5,5)(5,5)(5,5)?

Answer: 252525. (5−1)2+(5−2)2=16+9=25(5-1)^2+(5-2)^2=16+9=25(5−1)2+(5−2)2=16+9=25 before taking square root.

Flashcard 30: Find the distance between (−4,−3)(-4,-3)(−4,−3) and (0,0)(0,0)(0,0).

Answer: 555. (−4)2+(−3)2=16+9=25=5\sqrt{(-4)^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5(−4)2+(−3)2​=16+9​=25​=5