Line k is defined by y = 3x + 15. Line j is perpendicular to line k in the xy-plane....

1. TRANSLATE the given information

• Given: Line k has equation \(\mathrm{y = 3x + 15}\)
• Given: Line j is perpendicular to line k
• Find: The slope of line j

2. TRANSLATE the slope from line k's equation

• The equation \(\mathrm{y = 3x + 15}\) is in slope-intercept form \(\mathrm{y = mx + b}\)
• This means the slope of line k is \(\mathrm{m = 3}\)

3. INFER the relationship for perpendicular lines

• When two lines are perpendicular, their slopes are opposite reciprocals
• If one line has slope m, the perpendicular line has slope \(\mathrm{-1/m}\)
• Since we need the slope of line j (perpendicular to k), we need the opposite reciprocal of 3

4. SIMPLIFY to find the opposite reciprocal

• The reciprocal of 3 is \(\mathrm{1/3}\)
• The opposite reciprocal is \(\mathrm{-1/3}\)
• Therefore, the slope of line j is \(\mathrm{-1/3}\)

Answer: A. \(\mathrm{-1/3}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Missing conceptual knowledge: Students may not remember or understand that perpendicular lines have opposite reciprocal slopes, not just negative slopes.

Without this key relationship, students might think perpendicular simply means "negative slope" and look for \(\mathrm{-3}\) among the choices. Since \(\mathrm{-3}\) isn't an option, this leads to confusion and guessing among the available negative fractions.

Second Most Common Error:

Weak INFER reasoning: Students might remember that perpendicular lines involve reciprocals but forget the "opposite" part, leading them to think the slope should be \(\mathrm{1/3}\) instead of \(\mathrm{-1/3}\).

Since \(\mathrm{1/3}\) isn't among the answer choices, this causes them to get stuck and randomly select from the negative fraction options.

The Bottom Line:

This problem tests whether students truly understand the perpendicular line relationship beyond just "slopes are related." The key insight is that perpendicular lines require both the reciprocal AND the opposite (negative) to be applied together.