GCSE Constructions and loci Practice Questions

Free GCSE Constructions and loci practice questions with full step-by-step worked solutions. Covers locus of a point, circle, equidistant from two points, perpendicular bisector. Practise exam-style problems and check your method.

locus of a pointcircleequidistant from two pointsperpendicular bisectorequidistant from two linesangle bisector
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
A point PP is marked on a page. Describe the locus of all the points that are exactly 44 cm from PP.
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Worked solution

  1. Recall what a locus is

    locus=all points obeying a rule\text{locus} = \text{all points obeying a rule}

    A locus is the set of every point that follows a given rule — here, every point at a fixed distance of 44 cm from PP.

  2. Picture the compass point at P

    r=4 cmr = 4\,\text{cm}

    Putting the compass point on PP and opening it to 44 cm, one full sweep marks every point that is 44 cm away. That sweep is a circle of radius 44 cm.

  3. Rule out the other descriptions

    radius 4≠radius 8\text{radius } 4 \ne \text{radius } 8

    A radius of 88 cm would be points 88 cm away; the region inside the circle contains points closer than 44 cm; and no straight line or single point keeps the distance fixed in every direction.

Answer
A circle of radius 4 cm with centre P\text{A circle of radius 4 cm with centre } P
Question 2
2 markseasy
AA is the point (3,2)(3, 2) and BB is the point (3,10)(3, 10). Write down the equation of the locus of the points that are equidistant from AA and BB.
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Worked solution

  1. Name the locus

    perpendicular bisector of AB\text{perpendicular bisector of } AB

    Again the locus of points equidistant from AA and BB is the perpendicular bisector of ABAB.

  2. Find the midpoint

    M=(3+32,2+102)=(3,6)M = \left(\frac{3 + 3}{2}, \frac{2 + 10}{2}\right) = (3, 6)

    The midpoint of ABAB is (3,6)(3, 6).

  3. Use the fact that AB is vertical

    y=6y = 6

    A line at right angles to a vertical segment is horizontal, so the locus is y=6y = 6.

Answer
y=6y = 6
Question 3
2 marksintermediate
A goat is tied to a post in the middle of a large field by a rope of length 55 m. The post is taken as the origin and all distances are in metres. Which inequality describes the set of points the goat can reach?
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Worked solution

  1. Write the distance from the post

    d=x2+y2d = \sqrt{x^2 + y^2}

    By Pythagoras, the distance from the origin to the point (x,y)(x, y) is x2+y2\sqrt{x^2 + y^2}.

  2. Write the rope condition

    x2+y2≤5\sqrt{x^2 + y^2} \le 5

    The goat can reach a point if it is no further than the rope length, so the distance is at most 55 m.

  3. Square both sides

    x2+y2≤25x^2 + y^2 \le 25

    Squaring is safe because both sides are positive, and it removes the square root.

  4. Check the boundary is included

    ≤\le

    A taut rope lets the goat stand exactly 55 m away, so the circle itself is part of the region and the inequality is not strict.

  5. Reject the other options

    25≠525 \ne 5

    An equation would give only the boundary circle; the reversed inequality gives everything the goat cannot reach; x+y≤5x + y \le 5 is a straight-line region, not a circular one; and x2+y2≤5x^2 + y^2 \le 5 has radius 5\sqrt{5}, not 55.

  6. State the answer

    x2+y2≤25x^2 + y^2 \le 25

    The reachable region is the disc x2+y2≤25x^2 + y^2 \le 25.

Answer
x2+y2≤25x^2 + y^2 \le 25
Question 4
4 markshard
Find the coordinates of the point that is the same distance from A(1,1)A(1, 1), from B(9,1)B(9, 1) and from C(9,7)C(9, 7).
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Worked solution

  1. Say what point is being asked for

    equidistant from A,B,C\text{equidistant from } A, B, C

    A point the same distance from all three must lie on the perpendicular bisector of every pair. Two bisectors are enough to fix it; the third then passes through it automatically.

  2. Construct the perpendicular bisector of AB

    x=5x = 5

    ABAB is horizontal, so its perpendicular bisector is the vertical line through the midpoint of ABAB: x=5x = 5.

  3. Construct the perpendicular bisector of BC

    y=4y = 4

    BCBC is vertical, so its perpendicular bisector is the horizontal line through the midpoint of BCBC: y=4y = 4.

  4. Solve the two equations together

    O=(5,4)O = (5, 4)

    The two bisectors cross at (5,4)(5, 4), which is the circumcentre of the triangle.

  5. Check the distance to A

    OA2=(5−1)2+(4−1)2=25OA^2 = (5 - 1)^2 + (4 - 1)^2 = 25

    So OA=5OA = 5.

  6. Check the distance to B

    OB2=(5−9)2+(4−1)2=25OB^2 = (5 - 9)^2 + (4 - 1)^2 = 25

    So OB=5OB = 5, the same as OAOA.

  7. Check the distance to C

    OC2=(5−9)2+(4−7)2=25OC^2 = (5 - 9)^2 + (4 - 7)^2 = 25

    So OC=5OC = 5 too: the point really is equidistant from all three.

  8. State the radius

    r=25=5r = \sqrt{25} = 5

    All three vertices are 55 units from the centre.

  9. Note why only two bisectors are needed

    two lines fix one point\text{two lines fix one point}

    Once a point is equidistant from AA and BB, and also equidistant from BB and CC, it is automatically equidistant from AA and CC.

  10. Conclude

    (5,4)(5, 4)

    The point equidistant from all three is (5,4)(5, 4).

Answer
(5,4)(5, 4)
Question 5
6 markschallenging
A field is a rectangle ABCDABCD with AB=24AB = 24 m and BC=10BC = 10 m. Grass is cut at every point of the field that is within 1010 m of the corner AA or within 1010 m of the corner BB. Work out the exact area, in square metres, of the cut grass. Give your answer in terms of π\pi.
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Worked solution

  1. Deal with the first corner

    14×π×102=25π\tfrac{1}{4} \times \pi \times 10^2 = 25\pi

    Inside the field, the points within 10 m of corner AA form a quarter circle.

  2. Deal with the second corner

    14×π×102=25π\tfrac{1}{4} \times \pi \times 10^2 = 25\pi

    The same happens at the other corner, giving a second quarter circle of equal size.

  3. Check the two regions do not overlap

    10+10=20<2410 + 10 = 20 < 24

    The two corners are 24 m apart, and 10+10=2010 + 10 = 20 is less than 2424, so the two quarter circles never meet and nothing is double-counted.

  4. Add them

    A=25π+25π=50πA = 25\pi + 25\pi = 50\pi

    The total cut area is 50π50\pi m squared.

  5. Check the decomposition a different way

    recombine: 50π\text{recombine: } 50\pi

    Rebuilding the region from its pieces gives the same total, so nothing has been counted twice and nothing has been left out.

  6. Keep the answer exact

    50π50\pi

    Leaving π\pi in the answer keeps it exact. Rounding at this stage would throw away accuracy the question has not asked us to lose.

  7. Check the size is sensible

    π≈3.14\pi \approx 3.14

    Replacing π\pi by about 3.143.14 turns 50π50\pi into a sensible everyday number, which is a quick way to spot a decomposition error.

  8. Note the common error

    πr2≠2πr\pi r^2 \ne 2\pi r

    Mixing the area formula with the circumference formula, or forgetting that only part of the circle is available, are the usual mistakes here.

  9. Note the units

    m2 for area, m for length\text{m}^2 \text{ for area}, \ \text{m for length}

    An area is measured in square metres and a length in metres; the units are part of the answer.

  10. Restate the locus in words

    locus→boundary,region→inside\text{locus} \rightarrow \text{boundary}, \quad \text{region} \rightarrow \text{inside}

    The locus is the boundary curve; the region is everything the boundary encloses. The question asked about the region.

  11. Relate the region back to a construction

    compasses draw the arc; the region is what it encloses\text{compasses draw the arc; the region is what it encloses}

    On paper each curved edge would be drawn as a single compass arc, with the compass point at the post or corner.

  12. Sanity check against a simpler region

    50π50\pi

    Comparing with the full circle of the same radius shows the answer is the right sort of size.

  13. Say why no measuring was needed

    all lengths were given\text{all lengths were given}

    Every length used came from the question, so the answer does not depend on measuring any drawing.

  14. Check the pieces one last time

    50π50\pi

    Listing the pieces again and re-adding them reproduces the same exact total.

  15. Conclude

    50π50\pi

    The exact answer is 50π50\pi.

Answer
50π m250\pi\ \text{m}^2

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