Printable · GCSE Foundation · ages 14-16
Geometrical problems on coordinate axes worksheet — GCSE Foundation
Fifteen questions on "geometrical problems on coordinate axes" — DfE statement G11. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Answer key: Geometrical problems on coordinate axes worksheet — GCSE Foundation
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- (c) 5 hours — The distance from the origin to (7, 24) is √(7² + 24²) = √(49 + 576) = √625 = 25 km. Travelling at 5 km per hour, the time taken is 25 ÷ 5 = 5 hours. 25 hours mistakes the distance itself for the time, forgetting to divide by the speed. 125 hours comes from multiplying the distance by the speed, 25 × 5, instead of dividing. 0.2 hours comes from dividing the speed by the distance, 5 ÷ 25, the wrong way round.
- (b) 7 — Method: each coordinate of a midpoint is the mean of the matching pair of coordinates, so the y-coordinate of the midpoint depends on the two y-coordinates alone. Working: the y-coordinates are 4 and 10, so the mean is (4 + 10) ÷ 2 = 14 ÷ 2 = 7. Answer: 7. The distractors: 5 comes from working out the x-coordinate of the midpoint, (2 + 8) ÷ 2 = 5, and writing that down in place of the y-coordinate the question asked for; 3 comes from halving the difference of the y-coordinates, (10 − 4) ÷ 2 = 3, which is half the vertical gap rather than a position; 14 comes from adding the two y-coordinates and forgetting to halve the total.
- (b) Neither, because both coordinates differ — A line segment is horizontal only when both points share the same y-coordinate, and vertical only when both points share the same x-coordinate. Here A has x-coordinate −4 and B has x-coordinate 2, which differ, and A has y-coordinate 3 and B has y-coordinate 6, which also differ, so the segment is neither horizontal nor vertical. Every point has a y-coordinate and an x-coordinate, so simply having one is not a reason for the line to be horizontal or vertical — both of those wrong reasons ignore that the coordinates must match, not just exist. The x-coordinates do increase from A to B, but an increasing x-coordinate on its own describes a slope, not a horizontal line.
- (c) 24 — PQ lies along the x-axis with length 6, and PR lies along the y-axis with length 8, meeting at a right angle at P, so QR = √(6² + 8²) = √100 = 10. The perimeter is 6 + 8 + 10 = 24. "14" adds only the two shorter sides, PQ and PR, and leaves out the hypotenuse QR completely. "28" comes from finding QR incorrectly as 6 + 8 = 14 instead of using Pythagoras' theorem, then adding 6 + 8 + 14. "48" comes from multiplying the two shorter sides, 6 × 8, instead of finding and adding all three sides of the triangle.
- (d) positive — Moving from A to B, the x-coordinate increases from 1 to 4 and the y-coordinate also increases from 1 to 7, so the line rises as it goes from left to right, which means the gradient is positive. "negative" would need y to decrease as x increases, which is not the case here. "zero" would need the y-coordinate to stay the same, but it changes from 1 to 7. "cannot be determined" is wrong because the coordinates of both points are known, so the direction of the line can always be found.
- (c) (4, 3) — Method: the midpoint of a segment is the mean of its two end points, so its x-coordinate is the mean of the two x-coordinates and its y-coordinate is the mean of the two y-coordinates. Working: for x, (1 + 7) ÷ 2 = 8 ÷ 2 = 4. For y, (3 + 3) ÷ 2 = 6 ÷ 2 = 3. The midpoint is therefore (4, 3). Answer: (4, 3). The distractors: (3, 3) comes from halving the difference of the x-coordinates, (7 − 1) ÷ 2 = 3, which measures half the distance instead of locating the point; (3.5, 3) comes from halving only the larger x-coordinate and leaving the smaller one out of the working; (4, 0) comes from averaging the x-coordinates correctly but then subtracting the y-coordinates, 3 − 3, rather than averaging them.
- (a) 2 — Gradient = (change in y) ÷ (change in x) = (11 − 3) ÷ (6 − 2) = 8 ÷ 4 = 2. "0.5" comes from dividing the change in x by the change in y the wrong way round: 4 ÷ 8. "8" is only the change in y, forgetting to divide by the change in x at all. "−2" comes from a sign error, as if the y-coordinate had decreased rather than increased.
- (a) 18 — PQ lies along the x-axis with length 9, and PR lies along the y-axis with length 4, and these two sides meet at right angles at P, so they can be used as the base and height of the triangle. Area = 1/2 × base × height = 1/2 × 9 × 4 = 18. 36 comes from multiplying the base and height but forgetting to halve the result. 13 comes from adding the two lengths, 9 + 4, instead of multiplying them. 26 comes from the perimeter-style calculation 2 × (9 + 4) instead of the triangle area formula.
- (c) (2, 4) and (6, 4) — Method: two points lie on the same horizontal line when they stand at the same height above the x-axis, and a point's height is given by its y-coordinate, the second number in the bracket. Working: compare the second numbers in each pair. For (3, 5) and (3, 8) they are 5 and 8, which differ. For (1, 2) and (2, 1) they are 2 and 1, which differ. For (0, 0) and (1, 1) they are 0 and 1, which differ. For (2, 4) and (6, 4) they are 4 and 4, which agree, so only that pair lies on a horizontal line. Answer: (2, 4) and (6, 4). The distractors: (3, 5) and (3, 8) comes from swapping horizontal for vertical, since equal x-coordinates place two points one above the other; (1, 2) and (2, 1) comes from looking for the same pair of numbers rather than for the same y-coordinate; (0, 0) and (1, 1) comes from reading 'horizontal' as simply 'in a straight line', which any two points are.
- (c) 14 units — Method: the perimeter of a rectangle is the distance all the way round its outside, 2 × (length + width), so the two side lengths must be found first; on a coordinate grid a side's length is the difference between the coordinates that change along it. Working: along AB, from (1, 1) to (4, 1), only x changes, so AB = 4 − 1 = 3. Along BC, from (4, 1) to (4, 5), only y changes, so BC = 5 − 1 = 4. Perimeter = 2 × (3 + 4) = 2 × 7 = 14. Answer: 14 units. The distractors: 18 units comes from reading the vertex numbers 4 and 5 as the side lengths instead of subtracting, giving 2 × (4 + 5); 12 units comes from working out the area, 3 × 4, in place of the perimeter; 7 units comes from adding one length to one width and stopping there, without doubling for the opposite pair of sides.
- (d) 10 — Method: the distance between two points is the hypotenuse of a right-angled triangle whose shorter sides are the horizontal and vertical gaps, so work out both gaps first, handling the negative coordinates carefully, and then apply Pythagoras' theorem. Working: the horizontal gap is 5 − (−3) = 5 + 3 = 8 and the vertical gap is 4 − (−2) = 4 + 2 = 6. Then d² = 8² + 6² = 64 + 36 = 100, so d = √100 = 10. Answer: 10. The distractors: 14 comes from adding the two gaps, 8 + 6, instead of adding their squares and taking the root; 100 comes from stopping at the sum of the squares and never taking the square root; 50 comes from reaching 100 correctly and then halving it instead of taking its square root, a candidate who has read the last step as “halve” rather than “root”.
- (b) second quadrant — The point (−3, 5) has a negative x-coordinate and a positive y-coordinate, and this combination lies in the second quadrant. The first quadrant needs both coordinates positive. The third quadrant needs both coordinates negative. The fourth quadrant needs a positive x-coordinate and a negative y-coordinate.
- (b) 60 units² — Method: the area of a triangle is half the base times the perpendicular height, so choose a side to act as the base and measure the perpendicular distance from the opposite vertex to it. Working: A(0, 0) and B(10, 0) both lie on the x-axis, so AB is horizontal and AB = 10 − 0 = 10. The perpendicular height is the distance of C from the x-axis, which is its y-coordinate, 12. Area = (10 × 12) ÷ 2 = 120 ÷ 2 = 60. Answer: 60 units². The distractors: 120 units² comes from multiplying base by height and forgetting to halve; 65 units² comes from using the slanting side AC, which is 13 long, as the height in place of the perpendicular distance 12; 30 units² comes from halving the base to 5 before multiplying and then halving the product as well, so the halving is done twice.
- (b) (1, 6) — Method: when a square is set square-on to the grid, so that its sides run parallel to the axes, each vertex shares its x-coordinate with one neighbour and its y-coordinate with the other, and the missing vertex then borrows one coordinate from each of the two vertices it is joined to; so the first job is to check from the given points that the sides really do run parallel to the axes. Working: A(1, 2) and B(5, 2) share y = 2, so AB is a horizontal side; B(5, 2) and C(5, 6) share x = 5, so BC is a vertical side, which confirms that this square lies square-on to the axes and that the rule may be used. In square ABCD the vertex D is joined to C and to A. DC must be horizontal like AB, so D takes the y-coordinate of C, which is 6; DA must be vertical like CB, so D takes the x-coordinate of A, which is 1. D is therefore (1, 6), and checking confirms every side is 4 long. Answer: (1, 6). The distractors: (1, 5) comes from lifting the first number out of each of A and C, pairing the x-coordinate of A with the x-coordinate of C; (6, 1) comes from finding the right two numbers but writing them the wrong way round, height before sideways position; (9, 6) comes from stepping a further 4 to the right from C instead of closing the square back to the column A stands in.
- (a) (5, 2) — Reflecting in the x-axis keeps the x-coordinate the same and changes the sign of the y-coordinate, so (5, −2) maps to (5, 2). (−5, −2) changes the sign of the x-coordinate instead, which is what happens when reflecting in the y-axis. (−5, 2) changes the sign of both coordinates, which is the result of a rotation of 180° about the origin, not a reflection in the x-axis. (2, 5) comes from dropping the minus sign and then swapping the two numbers round, so neither coordinate has actually been reflected.
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