Compass constructions and loci

(a) With the compass point on $B$, draw an arc crossing both arms of the angle.

1. Step 1: arc centred on $B$ crossing both arms, at $X$ and $Y$.
2. Step 2: equal arcs from $X$ and $Y$ that cross.
3. Step 3: join $B$ to the crossing point.

• B1 The arc centred on $B$.

(b) The crossing point must be the same distance from both points, so it lies on the line of symmetry of the angle.

1. Equal radii make the crossing point equidistant from $X$ and $Y$, and $BX = BY$, so the line from $B$ through it is a line of symmetry of the angle.

• C1 Linking equal radii to the crossing point being the same distance from both arm points (symmetry of the construction).

(c) An arc centred on $B$ crossing both arms at $X$ and $Y$; equal arcs from $X$ and $Y$ crossing at $Z$; the line $BZ$, with every arc left visible.

1. With the compass point on $B$, draw an arc that crosses $BA$ at $X$ and $BC$ at $Y$.
2. Keep the compasses at one radius. Draw an arc from $X$ and an arc from $Y$ so that they cross at $Z$, inside the angle.
3. Draw the line from $B$ through $Z$. Leave every arc visible. Check with a protractor: the two halves should each be about $35^\circ$.

• B1 Arcs of equal radius drawn from two points on the arms that are the same distance from $B$, crossing inside the angle.
• B1 The bisector drawn from $B$ through the crossing point, within $2^\circ$ of the true bisector, with all construction arcs shown.

(a) 5 units

1. The shortest distance is the perpendicular distance, straight down from $P$ to $(3, 2)$: $7 - 2 = 5$.

• B1 Correct answer: 5.

(b) $122^\circ$

1. An obtuse angle is more than $90^\circ$, so read the scale that gives $122^\circ$.

• B1 Correct answer: $122^\circ$.

(c) 128 m

1. Real length $= 6.4 \times 2000 = 12\,800$ cm.
2. There are 100 cm in 1 m.
   $$12\,800 \div 100 = 128$$

• M1 $6.4 \times 2000$.
• A1 Correct answer: 128 m.

(a) $(2,7)$

1. $$\frac{-4+8}{2}$$
2. $$2\times 2+3$$
3. Points equidistant from A and B lie on the perpendicular bisector of AB.
4. AB is horizontal and its midpoint has x = ($-4$ + 8) $\div$ 2 = 2, so the bisector is x = 2.
5. Substitute x = 2 into y = 2x + 3: y = 7, so P is (2, 7).

• P1 Establishing $\frac{-4+8}{2}$ or an equivalent valid method.
• P1 Establishing $2\times 2+3$ or an equivalent valid method.
• A1 Correct answer: $(2,7)$

(a) The constructed line crosses AB at 2 cm from each end and is perpendicular to AB. Equal-radius arcs from A and B intersect on both sides of AB.

1. Use the same compass radius, greater than half AB, at A and B.
2. Join the two intersections of the arcs with a straight line.
3. The constructed line crosses AB at 2 cm from each end and is perpendicular to AB. Equal-radius arcs from A and B intersect on both sides of AB.

   AB

• M1 Use the same compass radius, greater than half AB, at A and B.
• M1 Join the two intersections of the arcs with a straight line.
• A1 Correct answer: The constructed line crosses AB at 2 cm from each end and is perpendicular to AB. Equal-radius arcs from A and B intersect on both sides of AB.

(b) The line is the perpendicular bisector of AB, the locus of points the same distance from its endpoints.

1. The line is the perpendicular bisector of AB, the locus of points the same distance from its endpoints.

• C1 Correct conclusion with supporting reasoning: The line is the perpendicular bisector of AB, the locus of points the same distance from its endpoints.

(a) Construct the internal angle bisector with equal arcs. Draw an arc centred at O of radius 8 cm. Its intersection with the bisector inside the angle is P.

1. Equal distances from the two rays locate P on the internal angle bisector.
2. The distance OP places P on a circle centred at O.
3. Construct the internal angle bisector with equal arcs. Draw an arc centred at O of radius 8 cm. Its intersection with the bisector inside the angle is P.

• M1 Equal distances from the two rays locate P on the internal angle bisector.
• M1 The distance OP places P on a circle centred at O.
• C1 Correct conclusion with supporting reasoning: Construct the internal angle bisector with equal arcs. Draw an arc centred at O of radius 8 cm. Its intersection with the bisector inside the angle is P.

(b) $45$°

1. Halve the right angle.
   $$90/2$$
2. Therefore $45$°.

• M1 Halve the right angle.
• A1 Correct answer: $45$°

(a) A circle of radius 5 m centred on the post, including every point inside it.

1. Every point within 5 m of the post is in reach: a circle of radius 5 m centred on the post, including every point inside it.

• C1 Correct conclusion with supporting reasoning: A circle of radius 5 m centred on the post, including every point inside it.

(b) $8$ m

1. The fence cuts the 5 m circle. The shortest line from the post to the fence is 3 m and meets it at right angles, so use Pythagoras with the rope as hypotenuse for half the reachable fence.
   $$\sqrt{5^{2}-3^{2}}=4$$
2. The goat reaches 4 m of fence on each side of that point.
   $$2\times 4=8$$
3. So the goat can reach $8$ m of fence.

• M1 Half the reachable fence: $\sqrt{5^2-3^2}=4$ m.
• A1 Correct answer: $8$ m

(a) $2$

1. Therefore $2$.

• B1 Correct answer: $2$

(b) $3$ cm

1. P lies above the midpoint of AB; use Pythagoras with half of AB.
   $$\sqrt{5^{2}-4^{2}}$$
2. Therefore $3$ cm.

• M1 P lies above the midpoint of AB; use Pythagoras with half of AB.
• A1 Correct answer: $3$ cm

(c) The perpendicular bisector of AB.

1. The perpendicular bisector of AB.

• C1 Correct conclusion with supporting reasoning: The perpendicular bisector of AB.

(a) $(1, 1)$

1. The midpoint of $AB$ is $(1, 1)$ and $AB$ is horizontal, so the perpendicular bisector is the vertical line $x = 1$.
2. The angle between the positive axes is $90^\circ$; its bisector makes $45^\circ$ with each axis: the line $y = x$.
3. Where $x = 1$ meets $y = x$: $P = (1, 1)$.

• P1 The perpendicular bisector $x = 1$.
• P1 The angle bisector $y = x$.
• A1 Correct answer: $(1, 1)$.

(b) 6 units

1. The shortest distance to a horizontal line is vertical: $7 - 1 = 6$.

• B1 Correct answer: 6.

(c) Any other route is the hypotenuse of a right-angled triangle with the perpendicular as a shorter side.

1. Join $P$ to any other point $Q$ on $y = 7$. Triangle $P$, $(1, 7)$, $Q$ has a right angle at $(1, 7)$, so $PQ$ is its hypotenuse, the longest side.

• C1 A correct argument: any slanted route is the hypotenuse of a right-angled triangle, so it is longer than the perpendicular.