In the coordinate plane, line segment AB connects points \(\mathrm{A(1, 3)}\) and \(\mathrm{B(7, 6)}\). A perpendicular line passes through point...

1. TRANSLATE the problem information

• Given information:
  • Line segment AB connects \(\mathrm{A(1, 3)}\) and \(\mathrm{B(7, 6)}\)
  • Need slope of line perpendicular to AB
  • The perpendicular line passes through \(\mathrm{C(4, 1)}\) (this point doesn't affect the slope)

2. INFER the solution strategy

• To find a perpendicular line's slope, I first need the original line's slope
• Then I'll use the perpendicular slope relationship

3. SIMPLIFY to find the slope of line AB

• Use slope formula: \(\mathrm{m = \frac{y_2 - y_1}{x_2 - x_1}}\)
• Slope of AB = \(\mathrm{\frac{6 - 3}{7 - 1}}\) = \(\mathrm{\frac{3}{6}}\) = \(\mathrm{\frac{1}{2}}\)

4. INFER the perpendicular slope

• Perpendicular lines have slopes that are negative reciprocals
• If original slope = \(\mathrm{\frac{1}{2}}\), then perpendicular slope = \(\mathrm{-\frac{1}{\frac{1}{2}}}\)

5. SIMPLIFY the negative reciprocal

• \(\mathrm{-\frac{1}{\frac{1}{2}}}\) = \(\mathrm{-1 \times \frac{2}{1}}\) = \(\mathrm{-2}}\)

Answer: D) \(\mathrm{-2}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students forget to make the slope negative when finding the perpendicular slope. They correctly find that AB has slope \(\mathrm{\frac{1}{2}}\), then take just the reciprocal (getting \(\mathrm{2}\)) instead of the negative reciprocal. This leads them to select Choice C (\(\mathrm{2}\)).

Second Most Common Error:

Poor SIMPLIFY execution: Students flip the coordinates incorrectly when calculating the original slope, getting \(\mathrm{\frac{7-1}{6-3}}\) = \(\mathrm{\frac{6}{3}}\) = \(\mathrm{2}\) instead of \(\mathrm{\frac{1}{2}}\). Then when they correctly apply the negative reciprocal rule, they get \(\mathrm{-\frac{1}{2}}\). This leads them to select Choice A (\(\mathrm{-\frac{1}{2}}\)).

The Bottom Line:

This problem tests whether students truly understand that perpendicular slopes are negative reciprocals, not just reciprocals. The key insight is that "perpendicular" always involves both flipping the fraction AND changing the sign.