In the coordinate plane, point A has coordinates \((-3, 8)\) and point B has coordinates \((11, -2)\). What is the...

1. TRANSLATE the problem information

• Given information:
  • Point A has coordinates \((-3, 8)\)
  • Point B has coordinates \((11, -2)\)
  • Need to find distance between them

• This tells us we need the distance formula for two points in the coordinate plane

2. INFER the approach

• We'll use the distance formula: \(\mathrm{d} = \sqrt{(\mathrm{x_2} - \mathrm{x_1})^2 + (\mathrm{y_2} - \mathrm{y_1})^2}\)
• This formula comes from the Pythagorean theorem applied to the right triangle formed by horizontal and vertical distances

3. TRANSLATE coordinates into the formula

• Let \(\mathrm{A}(-3, 8)\) be \((\mathrm{x_1}, \mathrm{y_1})\) and \(\mathrm{B}(11, -2)\) be \((\mathrm{x_2}, \mathrm{y_2})\)
• Calculate the differences:
  • \(\mathrm{x_2} - \mathrm{x_1} = 11 - (-3) = 11 + 3 = 14\)
  • \(\mathrm{y_2} - \mathrm{y_1} = -2 - 8 = -10\)

4. Apply the distance formula

• \(\mathrm{d} = \sqrt{(14)^2 + (-10)^2}\)
• \(\mathrm{d} = \sqrt{196 + 100}\)
  \(\mathrm{d} = \sqrt{296}\)

5. SIMPLIFY the radical

• Factor 296 to find perfect squares: \(296 = 4 \times 74\)
• Therefore: \(\sqrt{296} = \sqrt{4 \times 74}\)
  \(\sqrt{296} = 2\sqrt{74}\)

Answer: B. \(2\sqrt{74}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skill: Students correctly apply the distance formula and get \(\sqrt{296}\), but then either leave it unsimplified or factor incorrectly.

They might think \(\sqrt{296} = \sqrt{300} \approx 17.3\) and look for the closest numerical answer, leading them to select Choice C (24) as it seems like a reasonable distance.

Second Most Common Error:

Poor TRANSLATE reasoning: Students confuse distance formula with Manhattan distance (adding absolute differences instead of using Pythagorean theorem).

They calculate \(|11 - (-3)| + |-2 - 8| = 14 + 10 = 24\), leading them to select Choice C (24).

The Bottom Line:

This problem tests both formula application and radical simplification. Students who know the distance formula but struggle with factoring radicals will get stuck at the final step, while those who misunderstand what "distance" means geometrically may use incorrect methods entirely.