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GCSE · Loci and Construction

4.15D Loci and Construction: Fluency and Exam Drill

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 45 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Write down the locus of points that are exactly 3 cm from point A.

(Total for Question 1 is 1 mark)

Write down the locus of points that are equidistant from two distinct points A and B.

(Total for Question 2 is 1 mark)

Write down the locus of points that are exactly 4 cm from a given straight line l.

(Total for Question 3 is 1 mark)

Write down the locus of points equidistant from two intersecting straight lines.

(Total for Question 4 is 1 mark)

What is the locus of points at a distance less than or equal to 5 cm from point P?

(Total for Question 5 is 1 mark)

State a simple compass-and-straightedge construction to find the midpoint of segment AB.

(Total for Question 6 is 1 mark)

AB = 8 cm. M is the midpoint of AB. Work out the length AM.

(Total for Question 7 is 2 marks)

An angle measures 74 degrees. A student bisects it with ruler and compass. Work out the size of each angle after bisecting.

(Total for Question 8 is 2 marks)

Two towns S and T are 7 km apart. The region within 3 km of S and the region within 5 km of T are considered. Is there any point that lies in both regions? Give a reason.

(Total for Question 9 is 2 marks)

True or false? The locus of points equidistant from a fixed point F and a fixed line d is a parabola that can be constructed exactly using only compass and straightedge. Give a brief reason for your answer.

(Total for Question 10 is 2 marks)

Two circles have centres A and B which are 10 cm apart. One circle has radius 6 cm (centre A) and the other has radius 8 cm (centre B). Find the coordinates of their intersection points if AB lies along the x-axis with A at (0, 0) and B at (10, 0).

(Total for Question 11 is 3 marks)

Describe the steps to construct, with compass and straightedge, the perpendicular from a point P to a given line l (P not on l). Explain briefly why the construction produces a right angle.

(Total for Question 12 is 3 marks)

Triangle ABC has side lengths AB = 7 cm, BC = 8 cm and AC = 9 cm. Construct the perpendicular bisectors of two sides to find the circumcentre. Calculate the circumradius (to 3 significant figures).

(Total for Question 13 is 3 marks)

AB is a horizontal segment of length 10 cm with A at (0,0) and B at (10,0).
(a) Find the coordinates of the midpoint M of AB. (1 mark)
(b) Write down the equation of the perpendicular bisector of AB. (1 mark)
(c) A circle centre A has radius 6 cm. Find the intersection points of this circle with the perpendicular bisector. Give exact values where possible and decimal approximations to 3 s.f. (3 marks)

(a)Find the coordinates of the midpoint M of AB.(1)

(b)Write down the equation of the perpendicular bisector of AB.(1)

(c)Find the intersection points of circle centre A(0,0) radius 6 with the line x = 5. Give exact values where possible and decimal approximations to 3 s.f.(3)

(Total for Question 14 is 5 marks)

Two straight paths cross at right angles at O. A designer needs a point on the angle bisector between the two paths that is exactly 5 m from O. If we take the bisector as the line y = x, find the coordinates of the point in the first quadrant that is 5 m from O. Give your answer to 3 significant figures.

(Total for Question 15 is 4 marks)

A school S is at (0, 0), a hospital H is at (12, 0) and a landmark P is at (6, 8).
Find all points that are both equidistant from S and H and exactly 5 units from P. Give coordinates and justify your method. (8 marks)

(Total for Question 16 is 8 marks)

Mark scheme · 4.15D Loci and Construction: Fluency and Exam Drill

Question 1

B1 circle centre A radius 3 cm cao
Answer: A circle centre A radius 3 cm (the circle itself).

Question 2

B1 the perpendicular bisector of AB cao
Answer: The perpendicular bisector of AB.

Question 3

B1 two lines parallel to l, each 4 cm from l, cao
Answer: Two lines parallel to l, each 4 cm from l (one on each side).

Question 4

B1 the two angle bisectors of the lines (both bisectors) cao
Answer: The two angle bisectors of the intersecting lines.

Question 5

B1 the circle centre P radius 5 cm including the interior (a filled disc) cao
Answer: The circle centre P radius 5 cm together with its interior (the disc).

Question 6

B1 draw equal-radius arcs from A and B to make two intersection points; join intersections; intersection with AB is midpoint cao
Answer: With compass draw equal-radius arcs from A and B to make two intersections; join those intersections with a straight line; where that line meets AB is the midpoint.

Question 7

M1 take half of 8 or state AM = AB/2
A1 4 cm cao
Answer: 4 cm

Question 8

M1 divide 74 by 2
A1 37 degrees cao
Answer: 37 degrees and 37 degrees

Question 9

M1 use intersection test: radii sum 3 + 5 = 8 which is > 7 and |5 - 3| = 2 which is < 7 so circles overlap
A1 Yes, there are points in both regions (the two discs intersect) cao
Answer: Yes. Because 3 + 5 = 8 > 7 and |5 - 3| = 2 < 7, the two discs intersect so there are points in both regions.

Question 10

B1 false, because the parabola is a quadratic locus and cannot be constructed exactly with a finite sequence of compass-and-straightedge steps (oe)
B1 give reason: construction of a full parabola requires plotting many points or higher-order methods, not a single compass-and-straightedge procedure oe
Answer: False. The parabola is a continuous quadratic locus and cannot be constructed exactly by a finite straightedge-and-compass procedure; it is usually drawn by plotting many points.

Question 11

M1 use formula x = (r1² - r2² + d²)/(2d) or set up equations and subtract to find x
M1 substitute r1 = 6, r2 = 8, d = 10 to get x = (36 - 64 + 100)/20 = 72/20 = 3.6
A1 y = ±√6² - 3.6² = ±4.8 so intersections (3.6, 4.8) and (3.6, -4.8) cao
Answer: Intersection points: (3.6, 4.8) and (3.6, -4.8). (Checks: 3.6² + 4.8² = 12.96 + 23.04 = 36 = 6².)

Question 12

M1 construct two equal-radius arcs centred at P that meet line l at two points A and B (or from P draw arcs to meet l at A and B)
M1 from A and B draw equal-radius arcs on the same side of l to meet at X; join P to X or join A and B to get the perpendicular; method stated
A1 reason: triangles formed are congruent so the line is at equal distances and meets l at 90 degrees (or arcs give perpendicular bisector) cao
Answer: Steps: From P draw arcs to meet l at A and B. With centres A and B and a radius larger than AB/2 draw arcs that meet at X. Join P to X; PX is perpendicular to l. Reason: the construction makes congruent triangles so PX meets l at equal angles, giving a right angle.

Question 13

M1 use Heron to find area: s = 12, area = √12(12-7)(12-8)(12-9) = √720
M1 use circumradius formula R = abc/(4*area) with a=8, b=9, c=7
A1 R = 7*8*9/(4*√720) approx 4.70 cm awrt
Answer: Circumradius R approx 4.70 cm (to 3 s.f.). Computation: s = 12, area = √720 approx 26.8328, R = 504/(4*26.8328) approx 4.698 -> 4.70.

Question 14

(a) B1 midpoint M = (5, 0) cao
(a) Answer: M = (5, 0)
(b) B1 equation x = 5 cao
(b) Answer: x = 5
(c) M1 substitute x = 5 into x² + y² = 36 to get y² = 36 - 25 = 11
(c) M1 take square root to get y = ±√11
(c) A1 intersection points (5, √11) and (5, -√11) approx (5, 3.32) and (5, -3.32) cao
(c) Answer: Intersection points: (5, ±√11) approx (5, 3.32) and (5, -3.32).

Question 15

M1 substitute y = x into x² + y² = 5² so 2x² = 25
M1 solve x² = 12.5 so x = √12.5
A1 x approx 3.5355339 so to 3 s.f. 3.54
A1 point is (3.54, 3.54) cao
Answer: Point approx (3.54, 3.54). (Since x = √12.5 approx 3.5355, to 3 s.f. 3.54.)

Question 16

M1 identify locus equidistant from S and H is the perpendicular bisector of SH; since S(0,0) and H(12,0), bisector is x = 6
M1 parametrise points on bisector as (6, y)
M1 apply distance condition to P(6,8): distance = |y - 8| = 5, set up equation |y - 8| = 5
M1 solve |y - 8| = 5 to get y = 13 or y = 3
M1 state candidate points (6, 13) and (6, 3)
M1 check each point lies on perpendicular bisector and so is equidistant from S and H (show distance to S equals distance to H for one point)
A1 both points satisfy the condition; give coordinates (6, 13) and (6, 3) cao
A1 give brief construction statement: on the map draw perpendicular bisector of SH and mark points on it at distance 5 from P (using compass) cao
Answer: Points are (6, 13) and (6, 3). Reason: equidistant locus is x = 6; distance to P(6,8) is |y - 8| = 5 so y = 13 or 3; both points lie on x = 6.

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