307. The Perpendicular Product · when slopes multiply to –1

Try to break it

Drag A around the circle. Lines AB and CD stay perpendicular at every position. The slopes m₁ = tan(α) and m₂ = tan(θ) drift, but their product m₁ · m₂ stays locked at −1 — except when one line goes vertical (slope undefined), where the rule breaks down conceptually. That's the only escape from m₁ · m₂ = −1.

How you build it

Construct two perpendicular lines and check that their slopes multiply to −1.

• Using the Line tool, draw a line AB through the origin O at any slant. Count its rise over run on the grid — that is its slope m₁.
• With the Perp tool, click line AB at O and then to one side to build CD perpendicular to AB. Count its rise over run — slope m₂. The two slopes are negative reciprocals, so m₁ × m₂ = −1.

The proof, step by step

Prove that the product of the slopes of two perpendicular lines is −1.

• Let the inclination of line AB be α. By definition, its slope m₁ = tan α.
• Since AB ⊥ CD, the inclination of CD is θ = 90° + α.
• The slope of CD is m₂ = tan(90° + α) = –cot α.
• Multiplying the slopes: m₁ · m₂ = tan α · (–cot α) = tan α · (–1/tan α) = –1.

Worked example

In a Cartesian plane, line L₁ passes through (0, 3) and (4, 11). Line L₂ is perpendicular to L₁ and passes through (2, 5). What is the slope of L₂?

Slope of L₁ = (11-3)/(4-0) = 2. Since L₁ ⊥ L₂, m₁ · m₂ = –1 ⇒ 2 · m₂ = –1 ⇒ m₂ = –1/2.

• -1/2 — correct
• 1/2
• -2
• 2