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.

Calculate the side lengths: $$DE = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$$; $$DF = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$$; $$EF = \sqrt{(12-5)^2 + (5-12)^2} = \sqrt{7^2 + (-7)^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2}$$. Since DE = DF = 13, the triangle is isosceles. To check if it's also a right triangle: $$13^2 + (7\sqrt{2})^2 = 169 + 98 = 267 \neq 13^2 = 169$$, so it's not a right triangle.

To find the distance from A(2,1) to C(8,5), use the distance formula: $$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{(8-2)^2 + (5-1)^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52}$$. 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).

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: $$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$.

First, find the distance from $$S(-1, -3)$$ to $$T(7, 3)$$:

$$ST = \sqrt{(7-(-1))^2 + (3-(-3))^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$$ units

Next, find the distance from $$T(7, 3)$$ to $$U(15, 9)$$:

$$TU = \sqrt{(15-7)^2 + (9-3)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$$ units

The total distance is $$ST + TU = 10 + 10 = 20$$ units, which is answer choice C.

Let's examine why the other answers are incorrect. Choice A gives $$2\sqrt{50}$$, which equals $$2\sqrt{25 \cdot 2} = 10\sqrt{2} \approx 14.14$$—this might result from incorrectly combining the distances. Choice B shows $$10\sqrt{2}$$, which is approximately $$14.14$$—this could come from finding only one distance or making an algebraic error. Choice D gives $$\sqrt{200}$$, which equals $$\sqrt{100 \cdot 2} = 10\sqrt{2}$$, 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.

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 $$MNK$$ is isosceles with $$MK = NK$$, you need to calculate the distance from $$K(3,4)$$ to both $$M(0,8)$$ and $$N(6,0)$$. Using the distance formula $$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$:

For $$MK$$: $$\sqrt{(3-0)^2 + (4-8)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

For $$NK$$: $$\sqrt{(3-6)^2 + (4-0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

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 ($$\sqrt{41}$$ units) would result from incorrectly calculating one of the coordinate differences—possibly using $$(3-0)^2 + (4-8)^2$$ but making an arithmetic error like $$9 + 32 = 41$$. Choice C ($$\sqrt{25}$$ 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—$$\sqrt{25} = 5$$, not just $$\sqrt{25}$$. This ensures your answer matches the expected format and demonstrates full understanding of radical simplification.

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.

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 gives $$\sqrt{25}$$ without simplifying to 5. Choice C might result from miscalculating coordinate differences. Choice D might result from adding coordinate differences instead of using the distance formula.

Calculate each distance: $$AB = \sqrt{(4-(-2))^2 + (7-3)^2} = \sqrt{36+16} = \sqrt{52}$$; $$BC = \sqrt{(6-4)^2 + (-1-7)^2} = \sqrt{4+64} = \sqrt{68}$$; $$AC = \sqrt{(6-(-2))^2 + (-1-3)^2} = \sqrt{64+16} = \sqrt{80}$$. Since $$80 > 68 > 52$$, side AC is longest. Choice A and B give correct calculations but wrong conclusions. Choice D is incorrect since the sides have different lengths.

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.