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.

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, here simplified since points share y-coordinate. 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, but here Δy=0 so it's just |Δx|; for example, (1,2) to (4,2) would be √((3)²+0)=3, squaring eliminates signs. For points E(-1,-2) and F(5,-2), Δx=5-(-1)=6 and Δy=-2-(-2)=0, square to 36 and 0, add to 36, square root to 6. The correct distance is 6, as applying the formula gives √((5-(-1))²+( -2-(-2))²)=√(36+0)=6, matching choice C. Common errors include half distance (3, choice A), √(something small) like √6, negative from signs (-6, choice D), or ignoring absolute value. The process is: (1) identify E(-1,-2) and F(5,-2), (2) subtract Δx=6, Δy=0, (3) square to 36 and 0, (4) add to 36, (5) square root to 6, (6) verify it's equal to |Δx| since horizontal. Visualizing, points on horizontal line, no triangle needed, distance is just the x-difference absolute value; avoid negative distances or unnecessary calculations.

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 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.

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(0,0) and Q(3,5), Δx=3-0=3 and Δy=5-0=5, square each to get 9 and 25, add to 34, and take the square root to get √34 ≈5.8. The correct distance is √34 ≈5.8, as applying the formula gives √((3-0)²+(5-0)²)=√(9+25)=√34≈5.8, matching choice D. Common errors include taxicab distance by adding (3+5=8, choice A), wrong square root like √(9+16)=5 (but that's for different Δy), arithmetic errors like 9+25=34 but ≈5.7 or 5.8 mix-up, or confusing with √8≈2.8 (choice C). The process is: (1) identify coordinates P(0,0) and Q(3,5), (2) subtract to find Δx=3 and Δy=5, (3) square to 9 and 25, (4) add to 34, (5) square root ≈5.8, (6) verify it's longer than 3 and 5. Visualizing, plot the points, form the right triangle, and the hypotenuse is ≈5.8; avoid adding instead of Pythagorean or incorrect rounding.

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 C(-4,1) and D(2,-3), $\Delta x=2-(-4)=6$ and $\Delta y=-3-1=-4$, square each to get 36 and 16, add to 52, and take the square root to get $\sqrt{52}=2\sqrt{13}$. The correct distance is $\sqrt{52}=2\sqrt{13}$, as applying the formula gives $\sqrt{(2-(-4))^2 + (-3-1)^2}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}$, matching choice A. Common errors include wrong differences like $\Delta x=2+4=6$ but $\Delta y=3+1=4$ leading to $\sqrt{36+16}=$same, but mistakes like $\sqrt{36}=6$ (choice B, ignoring y), $\sqrt{20}=2\sqrt{5}$ (wrong add), or 10 (adding 6+4). The process is: (1) identify C(-4,1) and D(2,-3), (2) subtract $\Delta x=6$, $\Delta y=-4$, (3) square to 36 and 16, (4) add to 52, (5) square root to $2\sqrt{13}$, (6) verify longer than 6 and 4. Visualizing, plot with negative coordinates, form triangle legs 6 and 4, hypotenuse $2\sqrt{13}$; avoid sign errors or simplifying incorrectly.

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.

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.

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 (-4,0) and (2,-3), Δx=2-(-4)=6 and Δy=-3-0=-3, square each to get 36 and 9, add to 45, then square root to get √45. Thus, the correct distance is √45, which matches choice A. A common error is taxicab distance by adding 6+3=9 (choice C), miscalculating as √(36+(-3)²)=√45 but confusing with √27 (choice B) or 6 (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 6 and 3, and the hypotenuse is √45; mistakes include adding instead of using Pythagorean or forgetting to square negative differences.

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.