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 R(-3,-1) and S(4,3), Δx=4-(-3)=7 and Δy=3-(-1)=4, square each to get 49 and 16, add to 65, and take the square root to get √65 ≈8.1. The correct distance is √65 ≈8.1, as applying the formula gives √((4-(-3))²+(3-(-1))²)=√(49+16)=√65≈8.1, matching choice C. Common errors include wrong approximation (≈7.0, choice B), taxicab 7+4=11, or wrong radical like √(49+9?)=√58, √33≈5.7 (choice D, maybe Δy=3+1=4 but square wrong). The process is: (1) identify R(-3,-1) and S(4,3), (2) subtract Δx=7, Δy=4, (3) square to 49 and 16, (4) add to 65, (5) square root ≈8.1, (6) verify longer than 7 and 4. Visualizing, plot negative points, triangle legs 7 and 4, hypotenuse ≈8.1; avoid rounding errors or adding instead.

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 robot moving from P(0,0) to Q(3,5), we calculate: x₂-x₁ = 3-0 = 3 and y₂-y₁ = 5-0 = 5, then d = √(3² + 5²) = √(9 + 25) = √34 ≈ 5.8. The correct distance is √34 ≈ 5.8 units, using the formula d = √((3-0)² + (5-0)²) = √(9 + 25) = √34 ≈ 5.8. A common error would be taxicab distance (adding not squaring: 3+5=8), which gives choice C, or arithmetic error like √(9+25) = √16 = 4. Process: (1) identify coordinates (P: (0,0), Q: (3,5)), (2) subtract (3-0=3 and 5-0=5), (3) square differences (3²=9, 5²=25), (4) add squares (9+25=34), (5) square root (√34≈5.8), (6) verify reasonable (distance 5.8 is longer than both |Δx|=3 and |Δy|=5). Starting from origin simplifies calculations since x₁=y₁=0.

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.

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.

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.

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 (3,5), Δx=3-0=3 and Δy=5-0=5, square each to get 9 and 25, add to 34, then square root to get √34. Thus, the correct distance is √34, which matches choice A. A common error is taxicab distance by adding 3+5=8 (choice B), forgetting the square root to get 34 (choice D), or miscalculating as √(9+7)=√16=4 (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 3 and 5, and the hypotenuse is √34; mistakes include adding instead of using Pythagorean or algebraic errors like √(a²+b²)=a+b.

This question tests finding the distance between coordinate points using $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ derived from the Pythagorean theorem with horizontal and vertical legs forming a right triangle. The distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ comes from the Pythagorean theorem: points $(x_1,y_1)$ and $(x_2,y_2)$ with horizontal leg $|x_2-x_1|$ and vertical leg $|y_2-y_1|$ form a right triangle (third vertex at $(x_2,y_1)$ or $(x_1,y_2)$), distance is hypotenuse $d=\sqrt{(\Delta x)^2+(\Delta y)^2}$; for example, $(1,2)$ to $(4,6)$ has $\Delta x=3$, $\Delta y=4$, so $d=\sqrt{9+16}=\sqrt{25}=5$, and squaring eliminates signs, so subtraction order doesn't matter. For points $(-2,4)$ and $(5,1)$, $\Delta x=5-(-2)=7$ and $\Delta y=1-4=-3$, square each to get 49 and 9, add to 58, then square root to get $\sqrt{58} \approx 7.6$ when rounded to the nearest tenth. Thus, the correct distance is 7.6, which matches choice B. A common error is taxicab distance by adding 7+3=10 (choice A), arithmetic error like 49+9=58 but $\sqrt{58}\approx7.6$ mistaken as 7.0 (choice C) or confusing with $\sqrt{40}\approx6.3$ (choice D). The process is: (1) identify coordinates, (2) subtract to find $\Delta x$ and $\Delta y$, (3) square differences, (4) add squares, (5) take square root, and (6) verify the distance is longer than both $|\Delta x|$ and $|\Delta y|$. Visualizing helps: plot the points, imagine the right triangle with legs 7 and 3, and the hypotenuse is $\sqrt{58}\approx7.6$; mistakes include forgetting to square or rounding errors.

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 (-2,4) and (5,1), Δx=5-(-2)=7 and Δy=1-4=-3, square each to get 49 and 9, add to 58, then square root to get √58 ≈7.6 when rounded to the nearest tenth. Thus, the correct distance is 7.6, which matches choice B. A common error is taxicab distance by adding 7+3=10 (choice A), arithmetic error like 49+9=58 but √58≈7.6 mistaken as 7.0 (choice C) or confusing with √40≈6.3 (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 7 and 3, and the hypotenuse is √58≈7.6; mistakes include forgetting to square or rounding errors.

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.

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.