Find Distance Using Pythagorean Theorem - 8th Grade Math

Point P is at the origin $(0,0)$ and point Q is at $(8,6)$ . What is the distance from P to Q?
Answer: $\sqrt{100}=10$ 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.
Point $P$ is located at $(-3, 4)$ and point $Q$ is located at $(5, -2)$ . If point $R$ is the midpoint of segment $PQ$ , what is the distance from point $P$ to point $R$ ?
Answer: $5$ units First find midpoint R: $R = \left(\frac{-3+5}{2}, \frac{4+(-2)}{2}\right) = (1, 1)$ . Then find distance from P(-3,4) to R(1,1): $d = \sqrt{(1-(-3))^2 + (1-4)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ . Choice B is the full distance from P to Q. Choice C incorrectly doubles the correct answer. Choice D gives $\sqrt{100}$ from calculation errors.
A right triangle has vertices at $A(1, 2)$ , $B(1, 8)$ , and $C(9, 2)$ . The right angle is at vertex $A$ . What is the length of the hypotenuse of this triangle?
Answer: $10$ units 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: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ . Side AB: From A(1,2) to B(1,8) $AB = \sqrt{(1-1)^2 + (8-2)^2} = \sqrt{0 + 36} = 6$ units Side AC: From A(1,2) to C(9,2) $AC = \sqrt{(9-1)^2 + (2-2)^2} = \sqrt{64 + 0} = 8$ units Side BC (hypotenuse): From B(1,8) to C(9,2) $BC = \sqrt{(9-1)^2 + (2-8)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$ units Choice B is correct: the hypotenuse is 10 units long. Choice A ( $\sqrt{164}$ ) 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 ( $\sqrt{100}$ ) is mathematically equivalent to choice B since $\sqrt{100} = 10$ , 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.
On a coordinate plane, the points are $(-4,0)$ and $(2,-3)$ . What is the distance between them?
Answer: $\sqrt{45}$ The distance formula, $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ , comes from the Pythagorean theorem. For $(-4,0)$ and $(2,-3)$ : $\Delta x = 2-(-4)=6$ and $\Delta y = -3-0=-3$ . Squaring gives $36+9=45$ , so $d=\sqrt{45}$ . Choice C ( $\sqrt{27}$ ) comes from squaring the differences incorrectly. Choice D (6) is only the horizontal distance, ignoring the vertical change. Choice B (9) adds $|\Delta x|+|\Delta y|$ instead of using the Pythagorean theorem.
A student wants the distance between $T(-1,4)$ and $U(5,-4)$ . Which expression correctly represents the distance?
Answer: $\sqrt{(5-(-1))^2+(-4-4)^2}$ 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.
A student walks from $M(2,3)$ to $N(6,6)$ on a coordinate map of the school. What is the distance $MN$ (in units)? Give your answer in simplest form.
Answer: $5$ 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).
A student plots points $G(1,-4)$ and $H(6,2)$ on a coordinate plane. What is the distance between the points? Give an exact answer in simplest radical form.
Answer: $\sqrt{61}$ 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.
A game board is drawn on a coordinate plane. A token moves from $(0,0)$ to $(3,5)$ . What is the distance the token moved? Give the exact value.
Answer: $\sqrt{34}$ 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.
Which pair of points is farther apart? Pair 1: $E(1,1)$ and $F(7,1)$ Pair 2: $G(2,2)$ and $H(6,5)$
Answer: Pair 1 is farther apart. 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 Pair 1: E(1,1) to F(7,1), we have x₂-x₁ = 7-1 = 6 and y₂-y₁ = 1-1 = 0, so d₁ = √(6² + 0²) = √36 = 6; for Pair 2: G(2,2) to H(6,5), we have x₂-x₁ = 6-2 = 4 and y₂-y₁ = 5-2 = 3, so d₂ = √(4² + 3²) = √(16 + 9) = √25 = 5. Pair 1 has distance 6 units (horizontal line), while Pair 2 has distance 5 units (3-4-5 triangle), so Pair 1 is farther apart. Note that Pair 1 points have same y-coordinate (horizontal line), making distance simply |x₂-x₁| = 6. Process: calculate each distance separately using formula, then compare: d₁ = 6 > d₂ = 5. Visualizing helps: Pair 1 is horizontal segment length 6, Pair 2 forms 3-4-5 right triangle with hypotenuse 5.
On a coordinate plane, points $J(-1,2)$ and $K(4,4)$ are connected, forming the hypotenuse of a right triangle with legs parallel to the axes. What is the length of $\overline{JK}$ ? (Round to the nearest tenth.)
Answer: $\sqrt{29}\approx 5.4$ 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.
Point $E(-1,-2)$ and point $F(5,-2)$ lie on the same horizontal line. What is the distance $EF$ ?
Answer: $6$ 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.
On a coordinate plane, two students marked points $A(1,2)$ and $B(4,6)$ for a scavenger hunt. What is the distance from $A$ to $B$ (in units)?
Answer: $5$ 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).
Two points on a coordinate plane are $(-2,4)$ and $(5,1)$ . What is the distance between them? (Round to the nearest tenth.)
Answer: $7.6$ 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.
A game board is drawn on a coordinate plane. A token moves from $(0,0)$ to $(3,5)$ . What is the distance the token moved? Give the exact value.
Answer: $\sqrt{34}$ 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.
A student claims the distance between $( -3,2)$ and $(1,-4)$ is $|1-(-3)|+|-4-2|=4+6=10$ . What is the correct distance between the points?
Answer: $\sqrt{52}$ 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.
A student plots points $P(2,3)$ and $Q(6,6)$ on a coordinate plane. What is the distance $PQ$ ? (You may use $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ .) Give your answer in simplest form.
Answer: $5$ 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.
A circle has its center at $(3, -2)$ and passes through the point $(7, 1)$ . What is the radius of this circle?
Answer: $5$ units The radius is the distance from the center $(3,-2)$ to the point $(7,1)$ on the circle: $r=\sqrt{(7-3)^2+(1-(-2))^2}=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$ . 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, $7-3=4$ , without accounting for the vertical distance at all. Choice D would come from adding the coordinate differences, $4+3=7$ , instead of using the distance formula.
Three points form a triangle: $D(0, 0)$ , $E(5, 12)$ , and $F(12, 5)$ . Which statement about this triangle is correct?
Answer: Triangle DEF is isosceles, since two sides are equal in length. Calculate the side lengths using the distance formula: $DE=\sqrt{5^2+12^2}=\sqrt{169}=13$ , $DF=\sqrt{12^2+5^2}=\sqrt{169}=13$ , and $EF=\sqrt{(12-5)^2+(5-12)^2}=\sqrt{49+49}=7\sqrt2\approx9.9$ . Since $DE=DF=13$ but $EF$ is a different length, exactly two sides are equal, so the triangle is isosceles. Choice A is wrong because two of the three sides, $DE$ and $DF$ , are equal. Choice C is wrong because $EF$ is a different length than the other two sides. Choice D is wrong because the three points do form a valid triangle with positive area; they aren't collinear.
Points $A(-2,4)$ and $B(3,-1)$ are plotted on a coordinate plane. What is the distance $AB$ ? (Round to the nearest tenth.)
Answer: $\sqrt{50}\approx 7.1$ 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.
A point $K(2,5)$ is connected to a point $L(x,1)$ . If the distance $KL$ is $5$ , what are the possible values of $x$ ?
Answer: $x=-1$ or $x=5$ 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.

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