Constructions

Think about the two properties of the line: how it meets AB and where it crosses AB.

Compare the compass opening with half the length of AB.

A bisector divides the segment into two equal lengths.

An angle bisector divides an angle into two equal angles.

The compass opening stays fixed while this arc is drawn.

The next pair of arcs must locate a point the same distance from C and D.

1. Keep one compass opening for both of the next arcs. Draw an arc from C and another from D so that they cross inside the angle.
2. Join O to their intersection to complete the bisector.
3. The completed diagram shows the construction and its arcs.

The finished line must pass through P and through the intersection of the equal-radius arcs.

1. The first arc makes PC = PD. The next two arcs meet at E, where EC = ED.
2. Drawing PE gives the perpendicular at P. Keep the arcs visible when doing this construction on paper.
3. The completed diagram shows the construction and its arcs.

Use the original point outside the line and the intersection of the second pair of arcs.

1. The first arc centred at P makes PA = PB. The equal-radius arcs centred at A and B locate Q with QA = QB.
2. Both P and Q are equally distant from A and B, so PQ is their perpendicular bisector. It passes through P and is perpendicular to CD.
3. The completed diagram shows the construction and its arcs.

The paired arcs need to establish equal distances from A and B.

1. Use the same compass opening for the arcs centred at A and B. Make the opening large enough for the arcs to cross.
2. Their intersections will then be equally distant from A and B, so joining them gives the perpendicular bisector.
3. The completed diagram shows the construction and its arcs.

Each new side is a radius of one of the construction arcs.

1. AC = AB = 6 cm because C lies on the arc centred at A. BC = BA = 6 cm because C lies on the arc centred at B.
2. The triangle has three equal sides, so its perimeter is 6+6+6=18 cm.
3. The completed diagram shows the construction and its arcs.

First locate the midpoint of AB. Since AB is horizontal, its perpendicular bisector is vertical.

Part (a)

1. The midpoint x-coordinate is (-4+6)\div2=1. The y-coordinate remains 2.
2. M is (1, 2).
3. The completed diagram shows the construction and its arcs.

Part (b)

1. The perpendicular bisector is the vertical line through M, so every point on it has x-coordinate 1.
2. With y-coordinate 7, Q is (1, 7).
3. The completed diagram shows the construction and its arcs.

The two angles made by a bisector are equal.

1. 3x+5=5x-17.
2. Adding 17 and subtracting 3x gives 22=2x, so x=11.
3. Check: both angles are 38^\circ.
4. The completed diagram shows the construction and its arcs.

The required line has to satisfy both the angle condition and the point condition.

1. Line ℓ is the perpendicular bisector of the whole segment CD. It is perpendicular to CD, but it does not pass through P.
2. To start the correct construction, draw an arc centred at P to mark two points on CD, then construct their perpendicular bisector.
3. The completed diagram shows the construction and its arcs.

Consider how the first two points on the rays should be located.

1. The first points on the rays must be equally distant from O. Using a single arc centred at O makes these two distances equal.
2. The equal-radius arcs from those points can then be used to locate the angle bisector. Starting with OA = 3 cm and OB = 5 cm does not guarantee it.
3. The completed diagram shows the construction and its arcs.

Find the half of the right angle first, then use the straight line.

1. Angle CPB is 90° because PC is perpendicular to AB.
2. \angle DPB=90\div2=45^\circ.
3. \angle APD=180-45=135^\circ.
4. The completed diagram shows the construction and its arcs.

Find the midpoint of the vertical segment AB, then use the equation of the other line.

1. The midpoint of AB is (3,(-2+6)\div2)=(3,2).
2. AB is vertical, so its perpendicular bisector is the horizontal line y = 2.
3. On y=x+7, substitute y=2: 2=x+7, so x=-5.
4. P is (−5, 2).
5. The completed diagram shows the construction and its arcs.

The arcs must cross rather than just touch. Include both the rule about the arcs and the limit on the compasses.

1. The opening must be greater than half of 13 cm: 13\div2=6.5 cm.
2. The opening must also be a whole number no greater than 9 cm.
3. The possible openings are 7 cm, 8 cm and 9 cm.
4. The completed diagram shows the construction and its arcs.

A perpendicular bisector finds points equally distant from the two ends of a segment.

Part (a)

1. The perpendicular bisector of AB gives PA = PB. The perpendicular bisector of AC gives PA = PC.
2. At their intersection all three distances are equal.

Part (b)

1. The midpoint of AB is (3,0), so its perpendicular bisector is x=3.
2. The midpoint of AC is (-2,3), so its perpendicular bisector is y=3.
3. Their intersection is P = (3, 3).
4. The completed diagram shows the construction and its arcs.

The line y=-1 is horizontal, so a line perpendicular to it is vertical. The first arc is centred at P, so its two crossing points are the same distance from P.

Part (a)

1. The line y=-1 is horizontal, so the perpendicular to it is a vertical line.
2. The vertical line through P(6,5) is x=6. It meets y=-1 at R(6,-1).

Part (b)

1. C and D are both on the arc centred at P, so PC = PD. They lie evenly either side of the perpendicular x=6.
2. C is 6-2=4 units to the left of x=6, so D is 4 units to the right: x=6+4=10.
3. D is on the line y=-1, so D is (10,-1). Equal-radius arcs centred at C and D then meet on the line x=6, completing the perpendicular.

Use the compass-arc intersections to identify the perpendicular bisector. Then consider a right-angled triangle formed by half of one diagonal.

Part (a)

1. B and D are each equally distant from A and C. The line through both intersections is therefore the perpendicular bisector of AC.

Part (b)

1. AM=16\div2=8 cm. Since BD is perpendicular to AC, triangle AMB is right-angled.
2. AB=\sqrt{8^2+6^2}=\sqrt{100}=10 cm.
3. All four sides of a rhombus are equal, so its perimeter is 4 × 10 = 40 cm.
4. The completed diagram shows the construction and its arcs.