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.
Geometrical problems on coordinate axes worksheet — GCSE Foundation
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- 1.State whether the line segment joining A(−4, 3) and B(2, 6) is horizontal, vertical or neither, and give a reason for your answer.
- 2.Which pair of points lies on a horizontal line when the points are plotted on a coordinate grid?
- 3.A rectangle has vertices at (2, 1), (9, 1), (9, 5) and (2, 5). Work out the area of the rectangle.
- 4.The point (2, 5) is translated 3 units right and 4 units down. Work out the coordinates of the image point.
- 5.A path goes from A(1, 1) to B(1, 5), then from B to C(6, 5). Work out the total length of the path from A to C.
- 6.Work out the distance between the points (−3, 4) and (5, −2).
- 7.A line segment joins A(1, 1) and B(4, 7). Write down the sign of the gradient of this line.
- 8.The point (4, −3) is reflected in the y-axis. Work out the coordinates of the image point.
- 9.Work out the length of the straight line segment joining the points (2, −4) and (−3, −4).
- 10.A rectangle has vertices A(1, 1), B(4, 1), C(4, 5) and D(1, 5). Work out the perimeter of the rectangle.
- 11.Work out the gradient of the line segment joining A(2, 3) and B(6, 11).
- 12.The point (0, −5) lies on one of the two coordinate axes. Write down which axis this is.
- 13.A triangle has vertices A(1, 1), B(5, 1) and C(5, 4). Work out the lengths of AB, BC and CA, and write down whether triangle ABC is right-angled.
- 14.Triangle PQR has vertices P(0, 0), Q(6, 0) and R(0, 8). Work out the perimeter of the triangle.
- 15.Write down which quadrant contains the point (−3, 5) when plotted on a coordinate grid.
Answer key
- (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) (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.
- (b) 28 — The width of the rectangle is the difference in x-coordinates, 9 − 2 = 7, and the height is the difference in y-coordinates, 5 − 1 = 4. The area is width × height = 7 × 4 = 28. 22 comes from using the perimeter formula, 2 × (7 + 4), instead of the area formula. 35 comes from multiplying 7 by 5 instead of 4, misreading one of the y-coordinates. 63 comes from multiplying 9 by 7, using an x-coordinate instead of the height.
- (c) (5, 1) — Moving right increases the x-coordinate, and moving down decreases the y-coordinate, so (2, 5) becomes (2 + 3, 5 − 4) = (5, 1). (5, 9) comes from adding 4 to the y-coordinate instead of subtracting, as if the point moved up rather than down. (−1, 1) comes from subtracting 3 from the x-coordinate instead of adding, as if the point moved left rather than right. (2, 1) keeps the x-coordinate unchanged and only applies the vertical move, forgetting the horizontal move altogether.
- (a) 9 — AB is a vertical segment, since A and B share the x-coordinate 1, and its length is the difference in y-coordinates: 5 − 1 = 4. BC is a horizontal segment, since B and C share the y-coordinate 5, and its length is the difference in x-coordinates: 6 − 1 = 5. The total path length is 4 + 5 = 9. 20 comes from multiplying the two lengths, 4 × 5, instead of adding them. 5 is only the length of BC, forgetting to include AB. 4 is only the length of AB, forgetting to include BC.
- (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”.
- (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.
- (d) (−4, −3) — Reflecting in the y-axis keeps the y-coordinate the same and changes the sign of the x-coordinate, so (4, −3) maps to (−4, −3). "(4, 3)" changes the sign of the y-coordinate instead, which is what happens when reflecting in the x-axis. "(−4, 3)" changes the sign of both coordinates, which is the result of a rotation of 180° about the origin, not a reflection in the y-axis. "(3, −4)" swaps the two coordinates around instead of reflecting either of them.
- (b) 5 — Method: the two points have the same y-coordinate, so the segment joining them runs horizontally and its length is the gap between the two x-coordinates; a length is a distance, so it is never negative. Working: the x-coordinates are 2 and −3, so the gap is 2 − (−3) = 2 + 3 = 5. Both points have y = −4, so there is no vertical part to add on. Answer: 5. The distractors: 1 comes from dropping the minus sign on −3 and working out 3 − 2 instead; 0 comes from subtracting the y-coordinates, which are equal, in place of the x-coordinates; 6 comes from counting the grid lines from −3 across to 2 inclusive, which counts one more than the number of gaps between them.
- (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.
- (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.
- (b) y-axis — A point lies on the y-axis when its x-coordinate is 0; here the x-coordinate of (0, −5) is 0, so it lies on the y-axis. "x-axis" would need the y-coordinate to be 0 instead, which is not the case here. "the origin" is the single point (0, 0), not a whole axis, and this point is not (0, 0). "both axes" would only be true for the origin itself, where both coordinates are 0.
- (d) Yes, since 3² + 4² = 5² — AB is horizontal with length 5 − 1 = 4, BC is vertical with length 4 − 1 = 3, and CA = √(4² + 3²) = √25 = 5. Since the two shorter sides satisfy 3² + 4² = 5², the triangle is right-angled, with the right angle at B. "No, since 3 + 4 ≠ 5" wrongly tests Pythagoras' theorem by adding the sides instead of squaring them first. "No, since AB, BC and CA are not all equal" confuses a right-angled triangle with an equilateral one — a triangle does not need equal sides to have a right angle. "Yes, since 4² + 5² = 3²" reaches the correct conclusion but puts the longest side, 5, on the wrong side of the equation, as if it were one of the two shorter sides instead of the hypotenuse.
- (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.
- (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.
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