36 exam-style questions on Equations of perpendicular bisectors — topic 3.5 of Geometry and Trigonometry in the IB Mathematics: Applications and Interpretation syllabus — with answers and explanations. Samples below; the full set is free with an account.
Determine the coordinates of the midpoint of the segment joining A(2, 3) and B(8, 11).
Answer: (5, 7)
Midpoint = ((2+8)/2, (3+11)/2) = (5, 7). The perpendicular bisector passes through this point.
The line segment AB has gradient 2. Determine the gradient of the perpendicular bisector of AB.
Answer: -1/2
A perpendicular line has gradient equal to the negative reciprocal: −1/2. Choosing 2 (parallel) or 1/2 (reciprocal, wrong sign) are common errors.
Find the equation of the perpendicular bisector of the segment joining A(0, 0) and B(4, 8).
Answer: y = -1/2 x + 5
Midpoint = (2, 4); gradient of AB = 8/4 = 2, so bisector gradient = −1/2. y − 4 = −1/2(x − 2) ⇒ y = −1/2 x + 5.
Determine the equation of the perpendicular bisector of the segment joining P(-1, 3) and Q(5, 7).
Answer: y = -3/2 x + 8
Midpoint = (2, 5); gradient of PQ = (7−3)/(5−(−1)) = 4/6 = 2/3, so bisector gradient = −3/2. y − 5 = −3/2(x − 2) ⇒ y = −3/2 x + 3 + 5 = −3/2 x + 8.
Two shops are located at A(0, 4) and B(6, 0). A boundary is drawn so that customers use whichever shop is nearer. Determine the equation of this boundary.
Answer: y = 3/2 x - 5/2
The boundary is the perpendicular bisector of AB. Midpoint = (3, 2); gradient of AB = (0−4)/(6−0) = −2/3, so bisector gradient = 3/2. y − 2 = 3/2(x − 3) ⇒ y = 3/2 x − 5/2.
Find the equation of the perpendicular bisector of the segment joining A(-2, 5) and B(4, -1).
Answer: y = x + 1
Midpoint = (1, 2); gradient of AB = (−1−5)/(4−(−2)) = −6/6 = −1, so bisector gradient = 1. y − 2 = 1(x − 1) ⇒ y = x + 1.
That was 6 of 36 Equations of perpendicular bisectors questions.
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36 Equations of perpendicular bisectors questions — free with an account. Spaced repetition, streaks and full exam simulations included.
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