Loci and Construction: Practise Online

Either paper 2 marks

\(CD\) is a straight line \(9\) cm long.

The diagram shows a construction made on it with a pair of compasses and a straight edge. The construction arcs have been left in.

(a)
Write down the name of the construction that has been drawn.

(b)
Write down what is true about the distance from any point on the drawn line to \(C\) and to \(D\).
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(a) The perpendicular bisector of \(CD\)
(b) Every point on the line is the same distance from \(C\) as it is from \(D\)
The two arcs are swung with the same radius, one from \(C\) and one from \(D\), and the line is drawn through the two points where they cross.

That line is the perpendicular bisector of \(CD\): it cuts \(CD\) exactly in half, and it does so at a right angle.
Both arcs were drawn with the same radius, so each crossing is exactly as far from \(C\) as it is from \(D\). Every point on the line through them has that same property.

This is why the perpendicular bisector is the locus of points equidistant from \(C\) and \(D\) — and it is the reason the construction is worth doing rather than measuring.
Either paper 3 marks

\(RS\) is a straight line \(14\) cm long. The perpendicular bisector of \(RS\) has been constructed, and it crosses \(RS\) at \(M\).

(a)
Write down the length of \(RM\).

(b)
The point \(T\) lies on the bisector, and \(TR = 9\) cm.

Write down the length of \(TS\).
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(a) \(7\ \text{cm}\)
(b) \(9\ \text{cm}\)
A bisector cuts a line in half, so \(M\) is the midpoint of \(RS\).

\(14 \div 2 = 7\) cm.
Every point on the perpendicular bisector of \(RS\) is the same distance from \(R\) as it is from \(S\) — that is what the equal compass radii guarantee.

\(T\) is on the bisector, so \(TS = TR = 9\) cm.

No measuring and no Pythagoras: the locus does the work.
- (b) M1T is on the perpendicular bisector, so it is equidistant from \(R\) and \(S\)
- (b) A1T\(S\) = \(9\) cm
Either paper 1 mark
Towns A and B are shown on a scale diagram. Scale: 1 cm represents 50 miles. What does the shaded area represent? Tick one box.
☐ all the points at least 100 miles from A

☐ all the points within 100 miles of B

☐ all the points within 2 miles of A

☐ all the points at least 2 miles from B
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all the points within 100 miles of B
The shaded area is the inside of a circle about town B, so the wording is "within", not "at least".

The radius on the diagram is \(2\) cm and \(1\) cm represents \(50\) miles.

\(2 \times 50 = 100\) miles

The shaded area is all the points within 100 miles of B.
Either paper 2 marks

The point \(P\) is marked on the sheet.

Use a pair of compasses to construct the locus of points that are exactly 2 cm from \(P\).

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the whole circle with centre P and radius 2 cm

Every point exactly 2 cm from \(P\) lies on a circle with centre \(P\) and radius 2 cm.

1. Open the compass against the ruler on the sheet until the pencil is 2 cm from the point. 2. Put the compass point on \(P\). 3. Swing the pencil all the way round, back to where it started. A part of a circle is only part of the locus.
- M1an arc of radius 2 cm with centre \(P\)
- A1the whole circle, radius 2 cm, centre \(P\)
Either paper 3 marks

Here is angle \(XYZ\), which is \(30^\circ\), with a construction drawn on it. The construction arcs have been left in.

(a)
State the name of the construction that has been drawn.

(b)
The bisector meets the arc between the arms at \(W\).

Find the size of angle \(XYW\).
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(a) The bisector of angle \(XYZ\)
(b) \(15^\circ\)
An arc from the vertex \(Y\) cuts both arms. Two more arcs of equal radius, one from each of those cuts, meet between the arms, and the line from \(Y\) through that meeting point is drawn.

That is the angle bisector — it splits angle \(XYZ\) into two equal angles.
The bisector splits \(30^\circ\) into two equal parts.

\(30 \div 2 = 15\)

So angle \(XYW = 15^\circ\).
- (b) M1bisector splits \(30^\circ\) into two equal parts: \(30 \div 2\)
- (b) A1angle \(XYW = 15^\circ\)
Either paper 3 marks

Here is the locus of all the points that are exactly \(7\) cm from the point \(P\).

(a)
Write down the mathematical name of this locus.

(b)
Find the total length of the locus.

Give your answer in terms of \(\pi\).
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(a) A circle of radius \(7\) cm with its centre at \(P\)
(b) \(14\pi\ \text{cm}\)
Every point that is the same distance from a fixed point lies on a circle, and that fixed distance is the radius.

So the locus is a circle of radius \(7\) cm, centre \(P\).
The length of the locus is the circumference of the circle.

\(C = 2\pi r = 2 \times \pi \times 7 = 14\pi\) cm.
- (b) M1circumference \(2\pi r\) with \(r = 7\)
- (b) A1length \(14\pi\) cm
Either paper 2 marks

\(X\) is the point \((-3,\ -4)\) and \(Y\) is the point \((1,\ -4)\).

Here are four points.

\((0,\ 0) \qquad (-1,\ 0) \qquad (-2,\ -4) \qquad (0,\ -8)\)

Exactly one of them is the same distance from \(X\) as it is from \(Y\).

Write down that point.
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\((-1,\ 0)\)
The points that are the same distance from \(X\) as from \(Y\) form the perpendicular bisector of \(XY\).

\(X\) and \(Y\) are both at height \(y = -4\), so the line joining them is horizontal and its perpendicular bisector is the vertical line through the midpoint.

The midpoint is halfway between \(x = -3\) and \(x = 1\):

\((-3 + 1) \div 2 = -1\)

So the bisector is the line \(x = -1\), and the only one of the four points on it is \((-1,\ 0)\).

Notice you never need the distances themselves. Recognising the locus is the whole method.
- M1midpoint of \(XY\) found, \(x = -1\), or the perpendicular bisector is the vertical line through it
- A1point \((-1,\ 0)\) chosen
Either paper 2 marks

The points \(P\) and \(Q\) are 7 cm apart.

Use the compass and the straightedge to construct the locus of points that are the same distance from \(P\) as from \(Q\).

You must show your construction arcs.

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the perpendicular bisector of PQ: arcs of one radius from P and from Q that cross on both sides, and the line through the two crossings

The points the same distance from \(P\) and \(Q\) lie on the perpendicular bisector of \(PQ\).

1. Set the compass on the ruler to more than half of 7 cm, that is more than 3.5 cm. Here 5 cm works. 2. Put the compass point on \(P\) and draw an arc on each side of where the line \(PQ\) would be. 3. Keep the same width. Put the compass point on \(Q\) and draw arcs that cross the first two. 4. Rule a line through the two points where the arcs cross.

Leave the arcs showing. They are the working, and they are what earns the first mark.
- M1arcs of one radius, more than 3.5 cm, from \(P\) and from \(Q\) that cross on both sides
- A1a line ruled through both crossings, and no other line
Either paper 2 marks

The diagram shows two edges of a field, \(BA\) and \(BC\), which meet at the corner \(B\).

A path across the field starts at \(B\) and is always the same distance from \(BA\) as from \(BC\).

Use the compass and the straightedge to construct the path. Show your construction arcs.

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the bisector of the 100° angle ABC: an arc from B that cuts BA and BC, arcs of one radius from the two cuts that cross, and a line from B through the crossing

Every point the same distance from \(BA\) and \(BC\) lies on the bisector of angle \(ABC\).

Put the compass point on \(B\) and draw an arc that cuts both \(BA\) and \(BC\).

Keep one width for the next two arcs: one with the compass point where the first arc cuts \(BA\), one where it cuts \(BC\). Make them cross inside the angle.

Rule a line from \(B\) through the crossing. That line is the path.
- M1an arc with centre B that cuts both BA and BC, then arcs of one radius from those two points that cross
- A1a line ruled from B through the crossing of the arcs, and no other line
Either paper 2 marks

\(P\) is a point on the line \(AB\), with \(AP = 6\) cm and \(PB = 3\) cm.

Use the compass and the straightedge to construct an angle of \(90^\circ\) at \(P\).

You must show all your construction arcs.

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a line at right angles to AB through P, 6 cm from A: arcs from P that cut AB on each side of it, arcs of one radius from those two points that cross, and a line from P through the crossing

Put the compass point on \(P\) and, without changing its width, draw an arc that cuts \(AB\) on each side of \(P\). The two cuts are the same distance from \(P\).

Take the compass back to the ruler and open it wider. From each of the two cuts in turn, draw an arc on the same side of the line so that the two arcs cross.

Rule a line from \(P\) through the crossing. It is the perpendicular bisector of the two cuts, so it makes \(90^\circ\) with \(AB\) at \(P\). Leave all the arcs showing.
- M1arcs of one radius with centre P that cut AB on each side of P
- A1arcs of one radius from those two points that cross, and a line from P through the crossing, with no other line