The perpendicular bisector of the line segment joining P(2, 6) and Q(8, 2) has equation:

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Question

The perpendicular bisector of the line segment joining P(2, 6) and Q(8, 2) has equation:

Accepted Answer

3x − 2y = 7 — Midpoint of PQ: M = ((2+8)/2, (6+2)/2) = (5, 4). Gradient of PQ: m = (2−6)/(8−2) = −2/3. Perpendicular gradient: m⊥ = 3/2 (negative reciprocal). Equation through M(5,4): y − 4 = (3/2)(x − 5) → 2y − 8 = 3x − 15 → 3x − 2y = 7. Option A has incorrect constant. Option B has wrong gradient. Option D uses the gradient of PQ instead of the perpendicular.

Answer

2x − 3y = 1

Answer

3x − 2y = 11

Answer

2x + 3y = 19

Answer

3x − 2y = 7

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