In the coordinate plane, point A is located at the origin \((0, 0)\). Point B is located at coordinates \((15\sqrt{3},...

1. TRANSLATE the problem information

• Given information:
  • Point A is at the origin: \((0, 0)\)
  • Point B is at coordinates: \((15\sqrt{3}, 20\sqrt{3})\)
  • Need to find: distance from A to B

2. INFER the approach

• This is asking for the straight-line distance between two points in the coordinate plane
• We need the distance formula: \(\mathrm{d} = \sqrt{(\mathrm{x_2}-\mathrm{x_1})^2 + (\mathrm{y_2}-\mathrm{y_1})^2}\)

3. TRANSLATE coordinates into the distance formula

• Substitute the coordinates:
  • \((\mathrm{x_1}, \mathrm{y_1}) = (0, 0)\)
  • \((\mathrm{x_2}, \mathrm{y_2}) = (15\sqrt{3}, 20\sqrt{3})\)
• \(\mathrm{d} = \sqrt{(15\sqrt{3}-0)^2 + (20\sqrt{3}-0)^2}\)
• \(\mathrm{d} = \sqrt{(15\sqrt{3})^2 + (20\sqrt{3})^2}\)

4. SIMPLIFY the expression step by step

• Calculate each squared term:
  • \((15\sqrt{3})^2 = 15^2 \times (\sqrt{3})^2 = 225 \times 3 = 675\)
  • \((20\sqrt{3})^2 = 20^2 \times (\sqrt{3})^2 = 400 \times 3 = 1200\)
• Add them: \(\mathrm{d} = \sqrt{675 + 1200} = \sqrt{1875}\)

5. SIMPLIFY the radical

• Factor 1875: \(1875 = 625 \times 3 = 25^2 \times 3\)
• Therefore: \(\sqrt{1875} = \sqrt{25^2 \times 3} = 25\sqrt{3}\)

Answer: C (\(25\sqrt{3}\))

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't immediately recognize this as a distance problem requiring the distance formula. Instead, they might try to work with the coordinates in other ways or get confused about what operation to perform. This leads to confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly apply the distance formula but make arithmetic errors when calculating \((15\sqrt{3})^2\) and \((20\sqrt{3})^2\), or they get \(\sqrt{1875}\) but don't simplify it to \(25\sqrt{3}\). This may lead them to select Choice A (\(15\sqrt{3}\)) or Choice B (\(20\sqrt{3}\)) by mistakenly thinking one of the coordinate values is the answer.

The Bottom Line:

This problem tests whether students can connect coordinate geometry with the distance formula and then execute multi-step radical simplification. The key insight is recognizing that distance problems always use the Pythagorean relationship, even when coordinates involve radicals.