Constructions and Loci - Higher

A circle keeps one distance fixed.

1. Every point on the circumference is exactly one radius from the centre.
2. The locus is all points exactly 3 cm from P.

Consider distances from the two centres of the arcs.

1. The equal-radius arcs construct the perpendicular bisector of AB.
2. Every point on that line is equidistant from A and B.

The internal bisector divides the angle into two equal angles.

1. The two angles on either side of OC are equal.
2. Therefore angle COB=78\div2=39\degree.

The shortest distance from a point to a line is perpendicular to the line.

1. Segment PC is perpendicular to the line.
2. It gives the shortest distance, so the nearest point is C.

Points may lie on either side of the line.

1. On each side, a parallel line at perpendicular distance 2 cm gives the required points.
2. Measure the distance perpendicularly on both sides of the original line.

3. Both parallel lines are needed.

The equal-radius arcs must meet in two places.

1. Arcs of the same radius from A and B must cross twice, once on each side of AB. Joining the two crossing points gives the perpendicular bisector.
2. Two arcs from A and B only cross if the radius is more than half of AB, that is more than 8\div2=4 cm.
3. With radius 3 cm the arcs do not meet. With radius 4 cm they just touch at the midpoint of AB, which gives only one point.
4. The equal-radius arcs cross twice; join the two crossings.

5. With radius 5 cm the arcs cross twice, so drawing arcs of radius 5 cm from both A and B works.

The new point must be equidistant from the two points on the arms.

1. Draw equal-radius arcs centred at D and E so they meet inside the angle.
2. The equal arcs centred at D and E meet at P inside the angle.

3. Joining that intersection to B completes the internal angle bisector.

Both P and Q are equidistant from A and B.

1. Because PA=PB and QA=QB, both P and Q lie on the perpendicular bisector of AB.
2. Join P to Q to give the required perpendicular.

Combine the inside of the circle with the correct side of the bisector.

1. The point must be inside the circle and on the A side of the dashed line.
2. Only W meets both conditions.

Use both parallel lines, one on each side of the centre line.

1. Each dashed parallel meets the circle twice.
2. The two parallels therefore give 2+2=4 points.

Work out the angles in triangle ADE.

1. Angle DAB=90^\circ because ABCD is a rectangle, so the bisector makes \angle DAE=90^\circ\div2=45^\circ.
2. In triangle ADE, \angle ADE=90^\circ, so \angle AED=180^\circ-90^\circ-45^\circ=45^\circ.
3. Triangle ADE has two equal angles, so it is isosceles: DE=AD=8 cm.
4. DC=AB=12 cm, so EC=12-8=4 cm.

The perpendicular bisector halves the base, creating a right-angled triangle.

1. Since PA=PB, P lies on the perpendicular bisector, so AM=8\div2=4 cm.
2. In right-angled triangle AMP, PA^2=4^2+3^2=25.
3. Therefore PA=5 cm.

The joining segment is horizontal, so its perpendicular bisector is vertical.

1. The midpoint has x-coordinate (-2+8)\div2=3.
2. The perpendicular bisector is the vertical line x=3.

The outer boundary is excluded by the strict distance condition.

1. 'At least 2 cm from P' means on or outside the smaller circle. 'Less than 4 cm from P' means strictly inside the larger circle.
2. W is inside the smaller circle, so it is less than 2 cm from P. Y is outside the larger circle.
3. Z is on the larger circle, exactly 4 cm from P, so it is not less than 4 cm away.
4. X lies between the two circles, so X is the point.

Find the perpendicular map length before applying the scale.

1. The shortest route to the road is the perpendicular PF.
2. Pythagoras gives PF=\sqrt{5^2-3^2}=4 cm on the map.
3. The real distance is 4\times2.5=10 km.

First use the perpendicular bisector, then check the remaining distance.

1. The perpendicular bisector of AB is x=4, so only W and X remain.
2. For W, the distance from A is \sqrt{4^2+3^2}=5 m.
3. For X, the squared distance is 4^2+5^2=41, greater than 6^2=36.
4. Intersect the perpendicular bisector with the strict distance band.

5. Only W meets all the conditions.

R is equidistant from A and B, so it lies on the perpendicular bisector of AB.

1. R is the same distance from A as from B, so it lies on the perpendicular bisector of AB. The midpoint of AB is (3,0), so this is the line x=3.
2. Let R=(3,y). The right-angled triangle from A to (3,0) to R has base 3 and hypotenuse AR=5.
3. Pythagoras: y^2+3^2=5^2, so y^2=25-9=16.
4. R is above the x-axis, so y=4 and R=(3,4).

Intersect the perpendicular bisectors of two sides.

1. A point equidistant from A and B lies on the perpendicular bisector of AB. AB runs along the x-axis from 0 to 8, so this is the line x=4.
2. A point equidistant from A and C lies on the perpendicular bisector of AC. AC runs up the y-axis from 0 to 6, so this is the line y=3.
3. The point equidistant from all three vertices is where these lines cross: (4,3).
4. The two perpendicular bisectors meet at the required point.

5. Check: its distance from A is \sqrt{4^2+3^2}=5, and its distances from B and C are also \sqrt{4^2+3^2}=5.

The equal perpendicular distances form two sides of a right-angled triangle.

1. The distance from P to a road is measured along the perpendicular, so PF=PG=5 m.
2. OFPG has four right angles and PF=PG, so it is a square: OF=5 m.
3. Pythagoras in right-angled triangle OFP: OP^2=5^2+5^2=50.
4. OP=\sqrt{50}=\sqrt{25\times2}=5\sqrt2 m.

Check the distances from A and B first, then each candidate's distance from the top and bottom sides of the field.

1. All four candidates are exactly 4 m from A.
2. The perpendicular bisector of AB is x=8, so Z is not closer to A.
3. X and Y lie on horizontal sides, so their distances from those sides are zero.
4. Combine the radius condition, the distance band and the side of the perpendicular bisector.

5. W is closer to A and 4 m from each horizontal side, so W is suitable.