Middle School Math Quiz: Find Distance Using Pythagorean Theorem
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Find Distance Using Pythagorean TheoremQuestion 1 of 20

A rectangular park has vertices at A(2,1)A(2, 1), B(8,1)B(8, 1), C(8,5)C(8, 5), and D(2,5)D(2, 5). Maria walks from vertex AA directly to vertex CC along a diagonal path. What is the length of Maria's walk?

52\sqrt{52} units
40\sqrt{40} units
32\sqrt{32} units
26\sqrt{26} units
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Middle School Math Quiz

Middle School Math Quiz: Find Distance Using Pythagorean Theorem

Practice Find Distance Using Pythagorean Theorem in Middle School Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Find Distance Using Pythagorean Theorem, giving you a quick way to practice the rules, question types, and explanations that matter most for Middle School Math.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A rectangular park has vertices at A(2,1)A(2, 1), B(8,1)B(8, 1), C(8,5)C(8, 5), and D(2,5)D(2, 5). Maria walks from vertex AA directly to vertex CC along a diagonal path. What is the length of Maria's walk?

  1. 52\sqrt{52} units (correct answer)
  2. 40\sqrt{40} units
  3. 32\sqrt{32} units
  4. 26\sqrt{26} units
Explanation: To find the distance from A(2,1) to C(8,5), use the distance formula: d=(x2−x1)2+(y2−y1)2=(8−2)2+(5−1)2=62+42=36+16=52d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{(8-2)^2 + (5-1)^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52}. Choice B incorrectly uses coordinates (6,2) instead of the actual difference. Choice C represents the distance if one coordinate difference was miscalculated as 4 instead of 6. Choice D represents the distance from A to a point like (7,4).

Question 2

In a coordinate system, point MM is at (0,8)(0, 8) and point NN is at (6,0)(6, 0). A point KK is located such that it forms an isosceles triangle MNKMNK where MK=NKMK = NK. If KK is at (3,4)(3, 4), what is the length of the equal sides?

  1. 1010 units
  2. 41\sqrt{41} units
  3. 25\sqrt{25} units
  4. 55 units (correct answer)
Explanation: When you see an isosceles triangle problem in coordinate geometry, you need to use the distance formula to find the lengths of the sides and verify which two sides are equal. Since triangle MNKMNK is isosceles with MK=NKMK = NK, you need to calculate the distance from K(3,4)K(3,4) to both M(0,8)M(0,8) and N(6,0)N(6,0). Using the distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}: For MKMK: (3−0)2+(4−8)2=9+16=25=5\sqrt{(3-0)^2 + (4-8)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 For NKNK: (3−6)2+(4−0)2=9+16=25=5\sqrt{(3-6)^2 + (4-0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 Both equal sides have length 5 units, confirming the isosceles condition. Looking at the wrong answers: Choice A (10 units) is exactly double the correct answer, suggesting confusion between radius and diameter or perhaps adding instead of using the distance formula properly. Choice B (41\sqrt{41} units) would result from incorrectly calculating one of the coordinate differences—possibly using (3−0)2+(4−8)2(3-0)^2 + (4-8)^2 but making an arithmetic error like 9+32=419 + 32 = 41. Choice C (25\sqrt{25} units) shows the intermediate step before simplifying the square root; while mathematically equivalent to 5, it's not in simplest form. Always simplify square roots completely when possible—25=5\sqrt{25} = 5, not just 25\sqrt{25}. This ensures your answer matches the expected format and demonstrates full understanding of radical simplification.

Question 3

Points A(−2,4)A(-2,4) and B(3,−1)B(3,-1) are plotted on a coordinate plane. What is the distance ABAB? (Round to the nearest tenth.)

  1. 41≈6.4\sqrt{41}\approx 6.4
  2. 26≈5.1\sqrt{26}\approx 5.1
  3. 25=5\sqrt{25}=5
  4. 50≈7.1\sqrt{50}\approx 7.1 (correct answer)
Explanation: This problem tests finding distance between coordinate points using d=√((x₂-x₁)²+(y₂-y₁)²) derived from Pythagorean theorem with horizontal/vertical legs forming right triangle. Distance formula d=√((x₂-x₁)²+(y₂-y₁)²) comes from Pythagorean theorem: points (x₁,y₁) and (x₂,y₂) with horizontal leg |x₂-x₁| and vertical leg |y₂-y₁| form right triangle (third vertex at (x₂,y₁) or (x₁,y₂)), distance is hypotenuse d=√((Δx)²+(Δy)²). For points A(-2,4) and B(3,-1), we calculate: x₂-x₁ = 3-(-2) = 5 and y₂-y₁ = -1-4 = -5, then d = √(5² + (-5)²) = √(25 + 25) = √50 ≈ 7.1. The correct distance is √50 ≈ 7.1 units, using the formula d = √((3-(-2))² + (-1-4)²) = √(25 + 25) = √50 ≈ 7.1. A common error would be arithmetic mistakes with negative coordinates or calculating √50 = √25 = 5, which gives choice D. Process: (1) identify coordinates (A: (-2,4), B: (3,-1)), (2) subtract (3-(-2)=5 and -1-4=-5), (3) square differences (5²=25, (-5)²=25), (4) add squares (25+25=50), (5) square root (√50≈7.1), (6) verify reasonable (distance 7.1 is longer than both |Δx|=5 and |Δy|=5). Note that √50 = √(25×2) = 5√2 ≈ 7.1.

Question 4

A student plots points G(1,−4)G(1,-4) and H(6,2)H(6,2) on a coordinate plane. What is the distance between the points? Give an exact answer in simplest radical form.

  1. 61\sqrt{61} (correct answer)
  2. 11\sqrt{11}
  3. 25+36\sqrt{25}+\sqrt{36}
  4. 1111
Explanation: This question tests finding the distance between coordinate points using d=√((x₂-x₁)²+(y₂-y₁)²) derived from the Pythagorean theorem with horizontal and vertical legs forming a right triangle. The distance formula d=√((x₂-x₁)²+(y₂-y₁)²) comes from the Pythagorean theorem: points (x₁,y₁) and (x₂,y₂) with horizontal leg |x₂-x₁| and vertical leg |y₂-y₁| form a right triangle (third vertex at (x₂,y₁) or (x₁,y₂)), distance is hypotenuse d=√((Δx)²+(Δy)²); for example, (1,2) to (4,6) has Δx=3, Δy=4, so d=√(9+16)=√25=5, and squaring eliminates signs, so subtraction order doesn't matter. For points G(1,-4) and H(6,2), Δx=6-1=5 and Δy=2-(-4)=6, square each to get 25 and 36, add to 61, and take the square root to get √61. The correct distance is √61, as applying the formula gives √((6-1)²+(2-(-4))²)=√(25+36)=√61, matching choice A. Common errors include √(25+36- something)=√11 (choice B), adding 5+6=11 (choice C), or wrong like √25 + √36=5+6=11 (choice D). The process is: (1) identify G(1,-4) and H(6,2), (2) subtract Δx=5, Δy=6, (3) square to 25 and 36, (4) add to 61, (5) square root √61, (6) verify longer than 5 and 6. Visualizing, plot including negative y, form triangle legs 5 and 6, hypotenuse √61; avoid adding legs or algebraic mistakes like √(a²+b²)=a+b.

Question 5

A point K(2,5)K(2,5) is connected to a point L(x,1)L(x,1). If the distance KLKL is 55, what are the possible values of xx?

  1. x=2x=2 or x=7x=7
  2. x=−2x=-2 only
  3. x=−1x=-1 or x=5x=5 (correct answer)
  4. x=6x=6 only
Explanation: This question tests finding the distance between coordinate points using d=√((x₂-x₁)²+(y₂-y₁)²) derived from the Pythagorean theorem, here solving for x given distance 5. The distance formula d=√((x₂-x₁)²+(y₂-y₁)²) comes from the Pythagorean theorem: points (x₁,y₁) and (x₂,y₂) with legs |Δx| and |Δy|, hypotenuse d; for example, (1,2) to (4,6) has Δx=3, Δy=4, d=5; squaring both sides helps solve equations. For K(2,5) and L(x,1), set √((x-2)²+(1-5)²)=5, square both sides to (x-2)²+16=25, (x-2)²=9, x-2=±3, so x=5 or x=-1. The correct values are x=-1 or x=5, as applying the formula and solving gives those, matching choice D. Common errors include wrong Δy (1-5=-4, but squared 16), solving (x-2)²=25-16=9 correctly but wrong roots like x=6 (choice A) or x=2+4=6, x=2-4=-2 (choice B), or arithmetic like 25-16=9, ±√9=±3 but add wrong (choice C). The process is: (1) set up formula with unknown x, (2) square both sides to eliminate radical, (3) subtract known (Δy)²=16, (4) solve quadratic (x-2)²=9, (5) take ± square root, add to 2, (6) verify both give d=5. Visualizing, points at y=5 and y=1, vertical leg 4, so horizontal legs ±3 make hypotenuse 5 (3-4-5 triangle); avoid one-sided solutions or sign errors.

Question 6

A student wants the distance between T(−1,4)T(-1,4) and U(5,−4)U(5,-4). Which expression correctly represents the distance?

  1. (5−4)2+(−4−(−1))2\sqrt{(5-4)^2+(-4-(-1))^2}
  2. (5−(−1))2+(−4−4)2\sqrt{(5-(-1))^2+(-4-4)^2} (correct answer)
  3. (5−(−1))+(−4−4)(5-(-1))+(-4-4)
  4. (5−(−1))+(−4−4)\sqrt{(5-(-1))+( -4-4)}
Explanation: This question tests finding the distance between coordinate points using d=√((x₂-x₁)²+(y₂-y₁)²) derived from the Pythagorean theorem, by identifying the correct expression. The distance formula d=√((x₂-x₁)²+(y₂-y₁)²) comes from the Pythagorean theorem: points (x₁,y₁) and (x₂,y₂) with legs |Δx| and |Δy|, hypotenuse d; for example, (1,2) to (4,6) has Δx=3, Δy=4, d=√(9+16)=5; order doesn't matter due to squaring. For T(-1,4) and U(5,-4), the correct expression is √((5-(-1))²+(-4-4)²)=√(6²+(-8)²)=√(36+64)=√100=10. This matches choice A, as it correctly subtracts x and y differences inside squares, then sums under the square root. Common errors include adding differences without squares (choice B, taxicab), putting sum inside square root without squares (choice C, √(negative)), or wrong subtractions like 5-4=1, -4-(-1)=-3 (choice D, √(1+9)=√10). The process is: (1) identify points T(-1,4) and U(5,-4), (2) choose expression with correct Δx=5-(-1)=6 and Δy=-4-4=-8, (3) ensure squares on each, (4) sum inside radical, (5) square root, (6) verify computes to positive distance. Visualizing, ensure expression forms right triangle with legs 6 and 8, hypotenuse 10; avoid non-formula expressions or subtraction errors.

Question 7

Two points on a map grid are C(−4,1)C(-4,1) and D(2,−3)D(2,-3). What is the distance CDCD? (Leave your answer in simplest radical form.)

  1. 20=25\sqrt{20}=2\sqrt{5}
  2. 1010
  3. 36=6\sqrt{36}=6
  4. 52=213\sqrt{52}=2\sqrt{13} (correct answer)
Explanation: This question tests finding the distance between coordinate points using d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} derived from the Pythagorean theorem with horizontal and vertical legs forming a right triangle. The distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} comes from the Pythagorean theorem: points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) with horizontal leg ∣x2−x1∣|x_2 - x_1| and vertical leg ∣y2−y1∣|y_2 - y_1| form a right triangle (third vertex at (x2,y1)(x_2, y_1) or (x1,y2)(x_1, y_2)), distance is hypotenuse d=(Δx)2+(Δy)2d = \sqrt{(\Delta x)^2 + (\Delta y)^2}; for example, (1,2) to (4,6) has Δx=3\Delta x=3, Δy=4\Delta y=4, so d=9+16=25=5d=\sqrt{9+16}=\sqrt{25}=5, and squaring eliminates signs, so subtraction order doesn't matter. For points C(-4,1) and D(2,-3), Δx=2−(−4)=6\Delta x=2-(-4)=6 and Δy=−3−1=−4\Delta y=-3-1=-4, square each to get 36 and 16, add to 52, and take the square root to get 52=213\sqrt{52}=2\sqrt{13}. The correct distance is 52=213\sqrt{52}=2\sqrt{13}, as applying the formula gives (2−(−4))2+(−3−1)2=36+16=52=213\sqrt{(2-(-4))^2 + (-3-1)^2}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}, matching choice A. Common errors include wrong differences like Δx=2+4=6\Delta x=2+4=6 but Δy=3+1=4\Delta y=3+1=4 leading to 36+16=\sqrt{36+16}=same, but mistakes like 36=6\sqrt{36}=6 (choice B, ignoring y), 20=25\sqrt{20}=2\sqrt{5} (wrong add), or 10 (adding 6+4). The process is: (1) identify C(-4,1) and D(2,-3), (2) subtract Δx=6\Delta x=6, Δy=−4\Delta y=-4, (3) square to 36 and 16, (4) add to 52, (5) square root to 2132\sqrt{13}, (6) verify longer than 6 and 4. Visualizing, plot with negative coordinates, form triangle legs 6 and 4, hypotenuse 2132\sqrt{13}; avoid sign errors or simplifying incorrectly.

Question 8

On a coordinate plane, two students marked points A(1,2)A(1,2) and B(4,6)B(4,6) for a scavenger hunt. What is the distance from AA to BB (in units)?

  1. 77
  2. 2525
  3. 55 (correct answer)
  4. 24\sqrt{24}
Explanation: This problem tests finding distance between coordinate points using d=√((x₂-x₁)²+(y₂-y₁)²) derived from Pythagorean theorem with horizontal/vertical legs forming right triangle. Distance formula d=√((x₂-x₁)²+(y₂-y₁)²) comes from Pythagorean theorem: points (x₁,y₁) and (x₂,y₂) with horizontal leg |x₂-x₁| and vertical leg |y₂-y₁| form right triangle (third vertex at (x₂,y₁) or (x₁,y₂)), distance is hypotenuse d=√((Δx)²+(Δy)²). For points A(1,2) and B(4,6), we calculate: x₂-x₁ = 4-1 = 3 and y₂-y₁ = 6-2 = 4, then d = √(3² + 4²) = √(9 + 16) = √25 = 5. The correct distance is 5 units, using the formula d = √((4-1)² + (6-2)²) = √(9 + 16) = √25 = 5. A common error would be taxicab distance (adding not squaring: 3+4=7), which gives choice A, or forgetting the square root (d=25), which gives choice C. Process: (1) identify coordinates (A: (1,2), B: (4,6)), (2) subtract (4-1=3 and 6-2=4), (3) square differences (3²=9, 4²=16), (4) add squares (9+16=25), (5) square root (√25=5), (6) verify reasonable (distance 5 is longer than both |Δx|=3 and |Δy|=4). Visualizing: plot points, imagine right triangle with legs 3 and 4, hypotenuse 5 (classic 3-4-5 right triangle).

Question 9

Point P is at the origin (0,0)(0,0) and point Q is at (8,6)(8,6). What is the distance from P to Q?

  1. 28\sqrt{28}
  2. 1414
  3. 100100
  4. 100=10\sqrt{100}=10 (correct answer)
Explanation: This question tests finding the distance between coordinate points using d=√((x₂-x₁)²+(y₂-y₁)²) derived from the Pythagorean theorem with horizontal and vertical legs forming a right triangle. The distance formula d=√((x₂-x₁)²+(y₂-y₁)²) comes from the Pythagorean theorem: points (x₁,y₁) and (x₂,y₂) with horizontal leg |x₂-x₁| and vertical leg |y₂-y₁| form a right triangle (third vertex at (x₂,y₁) or (x₁,y₂)), distance is hypotenuse d=√((Δx)²+(Δy)²); for example, (1,2) to (4,6) has Δx=3, Δy=4, so d=√(9+16)=√25=5, and squaring eliminates signs, so subtraction order doesn't matter. For points (0,0) and (8,6), Δx=8-0=8 and Δy=6-0=6, square each to get 64 and 36, add to 100, then square root to get √100=10. Thus, the correct distance is 10, which matches choice B. A common error is taxicab distance by adding 8+6=14 (choice A), forgetting the square root to get 100 (choice D), or arithmetic error like √(64+36)=√100=10 but confusing with √(4+24)=√28 (choice C). The process is: (1) identify coordinates, (2) subtract to find Δx and Δy, (3) square differences, (4) add squares, (5) take square root, and (6) verify the distance is longer than both |Δx| and |Δy|. Visualizing helps: plot the points, imagine the right triangle with legs 8 and 6, and the hypotenuse is 10; mistakes include adding instead of using Pythagorean or algebraic errors like √(a²+b²)=a+b.

Question 10

On a coordinate plane, point AA is at (1,2)(1,2) and point BB is at (4,6)(4,6). Draw the right triangle using a horizontal segment and a vertical segment, then use the Pythagorean theorem to find the distance ABAB.

  1. 2525
  2. 7\sqrt{7}
  3. 77
  4. 55 (correct answer)
Explanation: This question tests finding the distance between coordinate points using d=√((x₂-x₁)²+(y₂-y₁)²) derived from the Pythagorean theorem with horizontal and vertical legs forming a right triangle. The distance formula d=√((x₂-x₁)²+(y₂-y₁)²) comes from the Pythagorean theorem: points (x₁,y₁) and (x₂,y₂) with horizontal leg |x₂-x₁| and vertical leg |y₂-y₁| form a right triangle (third vertex at (x₂,y₁) or (x₁,y₂)), distance is hypotenuse d=√((Δx)²+(Δy)²); for example, (1,2) to (4,6) has Δx=3, Δy=4, so d=√(9+16)=√25=5, and squaring eliminates signs, so subtraction order doesn't matter. For points A(1,2) and B(4,6), Δx=4-1=3 and Δy=6-2=4, square each to get 9 and 16, add to 25, and take the square root to get distance 5. The correct distance is 5, as applying the formula gives √((4-1)²+(6-2)²)=√(9+16)=√25=5, matching choice B. Common errors include taxicab distance by adding instead of squaring (3+4=7, choice A), forgetting the square root (25, choice C), or arithmetic mistakes like √(9+4)=√13 or others leading to √7 (choice D). The process is: (1) identify coordinates A(1,2) and B(4,6), (2) subtract to find Δx=3 and Δy=4 (order doesn't matter for squaring), (3) square differences to 9 and 16, (4) add to 25 under the radical, (5) square root to 5, (6) verify it's reasonable as 5 is longer than both 3 and 4. Visualizing, plot the points, imagine the right triangle with horizontal and vertical legs from A to B, and the hypotenuse is the direct distance of 5; avoid mistakes like adding legs instead of using Pythagorean or forgetting to square root.

Question 11

A circle has its center at (3,−2)(3, -2) and passes through the point (7,1)(7, 1). What is the radius of this circle?

  1. 2525 units
  2. 55 units (correct answer)
  3. 44 units
  4. 77 units
Explanation: The radius is the distance from the center (3,−2)(3,-2) to the point (7,1)(7,1) on the circle: r=(7−3)2+(1−(−2))2=42+32=16+9=25=5r=\sqrt{(7-3)^2+(1-(-2))^2}=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5. Choice A stops one step early, leaving the answer as the value under the square root instead of taking the square root. Choice C would come from using only the horizontal distance between the two points, 7−3=47-3=4, without accounting for the vertical distance at all. Choice D would come from adding the coordinate differences, 4+3=74+3=7, instead of using the distance formula.

Question 12

Point SS is located at (−1,−3)(-1, -3) and point TT is located at (7,3)(7, 3). Point UU is positioned so that ST=TUST = TU. If UU is at (15,9)(15, 9), what is the total distance traveled when walking from SS to TT to UU?

  1. 1010 units
  2. 1414 units
  3. 2020 units (correct answer)
  4. 1616 units
Explanation: When you see a problem asking for total distance traveled through multiple points, you need to find the distance between each consecutive pair of points and add them together. This requires using the distance formula: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}. First, find the distance from S(−1,−3)S(-1, -3) to T(7,3)T(7, 3):
ST=(7−(−1))2+(3−(−3))2=82+62=64+36=100=10ST = \sqrt{(7-(-1))^2 + (3-(-3))^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 units Next, find the distance from T(7,3)T(7, 3) to U(15,9)U(15, 9):
TU=(15−7)2+(9−3)2=82+62=64+36=100=10TU = \sqrt{(15-7)^2 + (9-3)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 units The total distance is ST+TU=10+10=20ST + TU = 10 + 10 = 20 units, which is answer choice C. Let's examine why the other answers are incorrect. Choice A gives 2502\sqrt{50}, which equals 225⋅2=102≈14.142\sqrt{25 \cdot 2} = 10\sqrt{2} \approx 14.14—this might result from incorrectly combining the distances. Choice B shows 10210\sqrt{2}, which is approximately 14.1414.14—this could come from finding only one distance or making an algebraic error. Choice D gives 200\sqrt{200}, which equals 100⋅2=102\sqrt{100 \cdot 2} = 10\sqrt{2}, the same value as choice B but in unsimplified form. Remember to always calculate each leg of the journey separately when finding total distance traveled, then add the results. Don't try to shortcut by finding the direct distance from start to end—that's displacement, not total distance traveled.

Question 13

On a coordinate plane, points J(−1,2)J(-1,2) and K(4,4)K(4,4) are connected, forming the hypotenuse of a right triangle with legs parallel to the axes. What is the length of JK‾\overline{JK}? (Round to the nearest tenth.)

  1. 9=3\sqrt{9}=3
  2. 29≈5.4\sqrt{29}\approx 5.4 (correct answer)
  3. 7.07.0
  4. 21≈4.6\sqrt{21}\approx 4.6
Explanation: This problem tests finding distance between coordinate points using d=√((x₂-x₁)²+(y₂-y₁)²) derived from Pythagorean theorem with horizontal/vertical legs forming right triangle. Distance formula d=√((x₂-x₁)²+(y₂-y₁)²) comes from Pythagorean theorem: points (x₁,y₁) and (x₂,y₂) with horizontal leg |x₂-x₁| and vertical leg |y₂-y₁| form right triangle (third vertex at (x₂,y₁) or (x₁,y₂)), distance is hypotenuse d=√((Δx)²+(Δy)²). For points J(-1,2) and K(4,4), we calculate: x₂-x₁ = 4-(-1) = 5 and y₂-y₁ = 4-2 = 2, then d = √(5² + 2²) = √(25 + 4) = √29 ≈ 5.4. The correct distance is √29 ≈ 5.4 units, using the formula d = √((4-(-1))² + (4-2)²) = √(25 + 4) = √29 ≈ 5.4. A common error would be taxicab distance (adding not squaring: 5+2=7), which gives choice D, or calculating √9 = 3 from some arithmetic error. Process: (1) identify coordinates (J: (-1,2), K: (4,4)), (2) subtract (4-(-1)=5 and 4-2=2), (3) square differences (5²=25, 2²=4), (4) add squares (25+4=29), (5) square root (√29≈5.4), (6) verify reasonable (distance 5.4 is longer than both |Δx|=5 and |Δy|=2). The problem explicitly mentions this forms a right triangle with legs parallel to axes, confirming our approach.

Question 14

Point PP is located at (−3,4)(-3, 4) and point QQ is located at (5,−2)(5, -2). If point RR is the midpoint of segment PQPQ, what is the distance from point PP to point RR?

  1. 55 units (correct answer)
  2. 50\sqrt{50} units
  3. 1010 units
  4. 100\sqrt{100} units
Explanation: First find midpoint R: R=(−3+52,4+(−2)2)=(1,1)R = \left(\frac{-3+5}{2}, \frac{4+(-2)}{2}\right) = (1, 1). Then find distance from P(-3,4) to R(1,1): d=(1−(−3))2+(1−4)2=42+(−3)2=16+9=25=5d = \sqrt{(1-(-3))^2 + (1-4)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5. Choice B is the full distance from P to Q. Choice C incorrectly doubles the correct answer. Choice D gives 100\sqrt{100} from calculation errors.

Question 15

A right triangle has vertices at A(1,2)A(1, 2), B(1,8)B(1, 8), and C(9,2)C(9, 2). The right angle is at vertex AA. What is the length of the hypotenuse of this triangle?

  1. 164\sqrt{164} units
  2. 1010 units (correct answer)
  3. 1414 units
  4. 100\sqrt{100} units
Explanation: When you see a right triangle problem with coordinates, you need to find the distances between vertices using the distance formula, then identify which side is the hypotenuse (the longest side, opposite the right angle). Since the right angle is at vertex A, the hypotenuse connects vertices B and C. Let's find all three side lengths using the distance formula: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}. Side AB: From A(1,2) to B(1,8) AB=(1−1)2+(8−2)2=0+36=6AB = \sqrt{(1-1)^2 + (8-2)^2} = \sqrt{0 + 36} = 6 units Side AC: From A(1,2) to C(9,2)
AC=(9−1)2+(2−2)2=64+0=8AC = \sqrt{(9-1)^2 + (2-2)^2} = \sqrt{64 + 0} = 8 units Side BC (hypotenuse): From B(1,8) to C(9,2) BC=(9−1)2+(2−8)2=64+36=100=10BC = \sqrt{(9-1)^2 + (2-8)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 units Choice B is correct: the hypotenuse is 10 units long. Choice A (164\sqrt{164}) likely comes from incorrectly adding coordinates or making calculation errors. Choice C (14 units) might result from adding the two legs instead of using the Pythagorean theorem. Choice D (100\sqrt{100}) is mathematically equivalent to choice B since 100=10\sqrt{100} = 10, but choice B gives the simplified form. Remember: in coordinate geometry problems involving right triangles, always use the distance formula to find side lengths, and the hypotenuse is always the side opposite the right angle. Double-check your arithmetic, especially when squaring negative numbers.

Question 16

A student claims the distance between (−3,2)( -3,2) and (1,−4)(1,-4) is ∣1−(−3)∣+∣−4−2∣=4+6=10|1-(-3)|+|-4-2|=4+6=10. What is the correct distance between the points?​

  1. 52\sqrt{52} (correct answer)
  2. 100=10\sqrt{100}=10
  3. 40\sqrt{40}
  4. 1010
Explanation: This question tests finding the distance between coordinate points using d=√((x₂-x₁)²+(y₂-y₁)²) derived from the Pythagorean theorem with horizontal and vertical legs forming a right triangle. The distance formula d=√((x₂-x₁)²+(y₂-y₁)²) comes from the Pythagorean theorem: points (x₁,y₁) and (x₂,y₂) with horizontal leg |x₂-x₁| and vertical leg |y₂-y₁| form a right triangle (third vertex at (x₂,y₁) or (x₁,y₂)), distance is hypotenuse d=√((Δx)²+(Δy)²); for example, (1,2) to (4,6) has Δx=3, Δy=4, so d=√(9+16)=√25=5, and squaring eliminates signs, so subtraction order doesn't matter. For points (-3,2) and (1,-4), Δx=1-(-3)=4 and Δy=-4-2=-6, square each to get 16 and 36, add to 52, then square root to get √52. Thus, the correct distance is √52, which matches choice B, correcting the student's taxicab error of 4+6=10 (choice A). A common error is taxicab by adding (choice A), confusing with √100=10 (choice C) or √40 (choice D). The process is: (1) identify coordinates, (2) subtract to find Δx and Δy, (3) square differences, (4) add squares, (5) take square root, and (6) verify the distance is longer than both |Δx| and |Δy|. Visualizing helps: plot the points, imagine the right triangle with legs 4 and 6, and the hypotenuse is √52; mistakes include adding instead of using Pythagorean (taxicab vs Euclidean) or forgetting the square root.

Question 17

A student plots points P(2,3)P(2,3) and Q(6,6)Q(6,6) on a coordinate plane. What is the distance PQPQ? (You may use d=(x2−x1)2+(y2−y1)2d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.)

  1. 7\sqrt{7}
  2. 77
  3. 25\sqrt{25}
  4. 55 (correct answer)
Explanation: This question tests finding the distance between coordinate points using d=√((x₂-x₁)²+(y₂-y₁)²) derived from the Pythagorean theorem with horizontal and vertical legs forming a right triangle. The distance formula d=√((x₂-x₁)²+(y₂-y₁)²) comes from the Pythagorean theorem: points (x₁,y₁) and (x₂,y₂) with horizontal leg |x₂-x₁| and vertical leg |y₂-y₁| form a right triangle (third vertex at (x₂,y₁) or (x₁,y₂)), distance is hypotenuse d=√((Δx)²+(Δy)²); for example, (1,2) to (4,6) has Δx=3, Δy=4, so d=√(9+16)=√25=5, and squaring eliminates signs, so subtraction order doesn't matter. For points P(2,3) and Q(6,6), Δx=6-2=4 and Δy=6-3=3, square each to get 16 and 9, add to 25, then square root to get distance 5. The correct distance is 5, as applying the formula gives d=√((6-2)²+(6-3)²)=√(16+9)=√25=5. Common errors include using taxicab distance by adding instead of squaring (4+3=7), wrong square root (√25=6), arithmetic error (16+9=26), forgetting square root (d=25), or algebraic mistake like √(a²+b²)=a+b. The process is: (1) identify coordinates (point 1: (2,3), point 2: (6,6)), (2) subtract (Δx=4 and Δy=3, order doesn't matter for squaring), (3) square differences (16 and 9), (4) add squares (25), (5) square root (5), (6) verify reasonable (distance 5 is longer than both |4| and |3|). Visualizing: plot points, imagine right triangle with horizontal and vertical legs from P to Q, hypotenuse connects them directly; mistakes include adding instead of using Pythagorean (4+3≠5, taxicab vs Euclidean), √(a²+b²)≠a+b error, forgetting square root (stopping at d²=25), or negative distance from sign errors.

Question 18

A student claims the distance between (−3,2)( -3,2) and (1,−4)(1,-4) is ∣1−(−3)∣+∣−4−2∣=4+6=10|1-(-3)|+|-4-2|=4+6=10. What is the correct distance between the points?

  1. 52\sqrt{52} (correct answer)
  2. 40\sqrt{40}
  3. 1010
  4. 100=10\sqrt{100}=10
Explanation: This question tests finding the distance between coordinate points using d=√((x₂-x₁)²+(y₂-y₁)²) derived from the Pythagorean theorem with horizontal and vertical legs forming a right triangle. The distance formula d=√((x₂-x₁)²+(y₂-y₁)²) comes from the Pythagorean theorem: points (x₁,y₁) and (x₂,y₂) with horizontal leg |x₂-x₁| and vertical leg |y₂-y₁| form a right triangle (third vertex at (x₂,y₁) or (x₁,y₂)), distance is hypotenuse d=√((Δx)²+(Δy)²); for example, (1,2) to (4,6) has Δx=3, Δy=4, so d=√(9+16)=√25=5, and squaring eliminates signs, so subtraction order doesn't matter. For points (-3,2) and (1,-4), Δx=1-(-3)=4 and Δy=-4-2=-6, square each to get 16 and 36, add to 52, then square root to get √52. Thus, the correct distance is √52, which matches choice B, correcting the student's taxicab error of 4+6=10 (choice A). A common error is taxicab by adding (choice A), confusing with √100=10 (choice C) or √40 (choice D). The process is: (1) identify coordinates, (2) subtract to find Δx and Δy, (3) square differences, (4) add squares, (5) take square root, and (6) verify the distance is longer than both |Δx| and |Δy|. Visualizing helps: plot the points, imagine the right triangle with legs 4 and 6, and the hypotenuse is √52; mistakes include adding instead of using Pythagorean (taxicab vs Euclidean) or forgetting the square root.

Question 19

Two points are R(6,8)R(6,8) and S(2,3)S(2,3). What is the distance between them? (Round to the nearest tenth.)

  1. 41≈6.4\sqrt{41}\approx 6.4 (correct answer)
  2. 32≈5.7\sqrt{32}\approx 5.7
  3. 9≈3.0\sqrt{9}\approx 3.0
  4. 9.09.0
Explanation: This problem tests finding distance between coordinate points using d=√((x₂-x₁)²+(y₂-y₁)²) derived from Pythagorean theorem with horizontal/vertical legs forming right triangle. Distance formula d=√((x₂-x₁)²+(y₂-y₁)²) comes from Pythagorean theorem: points (x₁,y₁) and (x₂,y₂) with horizontal leg |x₂-x₁| and vertical leg |y₂-y₁| form right triangle (third vertex at (x₂,y₁) or (x₁,y₂)), distance is hypotenuse d=√((Δx)²+(Δy)²). For points R(6,8) and S(2,3), we calculate: x₂-x₁ = 2-6 = -4 and y₂-y₁ = 3-8 = -5, then d = √((-4)² + (-5)²) = √(16 + 25) = √41 ≈ 6.4. The correct distance is √41 ≈ 6.4 units, using the formula d = √((2-6)² + (3-8)²) = √(16 + 25) = √41 ≈ 6.4. A common error would be taxicab distance (adding absolute values: 4+5=9), which gives choice D, or arithmetic error in adding squares. Process: (1) identify coordinates (R: (6,8), S: (2,3)), (2) subtract (2-6=-4 and 3-8=-5), (3) square differences ((-4)²=16, (-5)²=25), (4) add squares (16+25=41), (5) square root (√41≈6.4), (6) verify reasonable (distance 6.4 is longer than both |Δx|=4 and |Δy|=5). Squaring eliminates negative signs, so order of subtraction doesn't matter.

Question 20

A student walks from M(2,3)M(2,3) to N(6,6)N(6,6) on a coordinate map of the school. What is the distance MNMN (in units)?

  1. 7\sqrt{7}
  2. 77
  3. 55 (correct answer)
  4. 25\sqrt{25}
Explanation: This problem tests finding distance between coordinate points using d=√((x₂-x₁)²+(y₂-y₁)²) derived from Pythagorean theorem with horizontal/vertical legs forming right triangle. Distance formula d=√((x₂-x₁)²+(y₂-y₁)²) comes from Pythagorean theorem: points (x₁,y₁) and (x₂,y₂) with horizontal leg |x₂-x₁| and vertical leg |y₂-y₁| form right triangle (third vertex at (x₂,y₁) or (x₁,y₂)), distance is hypotenuse d=√((Δx)²+(Δy)²). For points M(2,3) and N(6,6), we calculate: x₂-x₁ = 6-2 = 4 and y₂-y₁ = 6-3 = 3, then d = √(4² + 3²) = √(16 + 9) = √25 = 5. The correct distance is 5 units, using the formula d = √((6-2)² + (6-3)²) = √(16 + 9) = √25 = 5. A common error would be taxicab distance (adding not squaring: 4+3=7), which gives choice D, or not taking the square root and leaving as √25, which gives choice B. Process: (1) identify coordinates (M: (2,3), N: (6,6)), (2) subtract (6-2=4 and 6-3=3), (3) square differences (4²=16, 3²=9), (4) add squares (16+9=25), (5) square root (√25=5), (6) verify reasonable (distance 5 is longer than both |Δx|=4 and |Δy|=3). This forms another 3-4-5 right triangle (with legs reversed from first problem).