Printable · GCSE Higher · ages 14-16
Geometrical problems on coordinate axes worksheet — GCSE Higher
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 Higher
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- 1.Line 1 passes through (0, 1) and (2, 5). Line 2 passes through (3, 2) and (5, 6). Work out the gradient of each line. Then write down whether the two lines are parallel.
- 2.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.
- 3.Work out the distance between the points (−3, 4) and (5, −2).
- 4.A rectangle has vertices A(1, 1), B(4, 1), C(4, 5) and D(1, 5). Work out the perimeter of the rectangle.
- 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.A vertical line passes through the point (5, 2). Write down the equation of this line.
- 7.Triangle PQR has vertices P(0, 0), Q(9, 0) and R(0, 4). Work out the area of triangle PQR.
- 8.Work out the y-coordinate of the midpoint of the line segment joining (2, 4) and (8, 10).
- 9.The point A(3, 2) is translated by the vector (−5, 4) to give point B. Work out the coordinates of B.
- 10.A(1, 2), B(5, 2) and C(5, 6) are three of the four vertices of a square ABCD. Work out the coordinates of D.
- 11.A is the point (1, 3) and B is the point (7, 3). Work out the coordinates of the midpoint of AB.
- 12.The midpoint of the line segment AB is (3, 5). A is the point (1, 3). Work out the coordinates of B.
- 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.A rectangle has vertices A(0, 0), B(6, 0), C(6, 4) and D(0, 4). Work out the length of the diagonal AC. Give your answer correct to 2 decimal places.
- 15.A triangle has vertices A(0, 0), B(10, 0) and C(5, 12). Work out the area of the triangle.
Answer key
- (a) Parallel, since both gradients are 2 — Gradient of line 1 = (5 − 1) ÷ (2 − 0) = 4 ÷ 2 = 2. Gradient of line 2 = (6 − 2) ÷ (5 − 3) = 4 ÷ 2 = 2. The two gradients are equal, so the lines are parallel. "Not parallel, since the gradients are 2 and 4" uses 4 for line 2, which comes from dividing the change in y, 4, by 1 rather than by the change in x, 5 − 3 = 2. "Not parallel, since the two lines cross the y-axis at different points" confuses parallel lines with the same line repeated — parallel lines are distinct lines, and where a line crosses the y-axis has no effect on its gradient. "Parallel, since both lines pass through positive coordinates" reaches the right conclusion for the wrong reason: lines are parallel because their gradients are equal, not because the coordinates given happen to be positive.
- (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.
- (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”.
- (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) 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.
- (b) x = 5 — Every point on a vertical line has the same x-coordinate, so the equation of a vertical line through (5, 2) is x = 5. "y = 5" mixes up the coordinates, using the x-value of 5 to write a y-equation. "y = 2" is the equation of the horizontal line through (5, 2), not the vertical one. "x = 2" uses the correct letter but the wrong coordinate, the y-value of 2 instead of the x-value of 5.
- (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.
- (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.
- (a) (−2, 6) — Translating by the vector (−5, 4) means adding −5 to the x-coordinate and adding 4 to the y-coordinate: (3 + (−5), 2 + 4) = (−2, 6). (8, 6) comes from treating −5 as +5, adding instead of subtracting on the x-coordinate. (−2, −2) keeps the x-coordinate correct but subtracts 4 from the y-coordinate instead of adding it. (7, −3) comes from swapping the two components of the vector, applying 4 to the x-coordinate and −5 to the y-coordinate.
- (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.
- (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) (5, 7) — Method: a midpoint is the mean of the two end points, so for each coordinate (start + end) ÷ 2 = midpoint; rearranging that gives end = 2 × midpoint less the start. Working: for x, (1 + x) ÷ 2 = 3, so 1 + x = 6 and x = 5. For y, (3 + y) ÷ 2 = 5, so 3 + y = 10 and y = 7. B is therefore (5, 7). Answer: (5, 7). The distractors: (2, 2) comes from subtracting A from the midpoint, (3 − 1, 5 − 3), which gives the step from A to the midpoint and stops there instead of taking that same step a second time; (4, 8) comes from adding A to the midpoint, (3 + 1, 5 + 3), without doubling the midpoint first; (6, 10) comes from doubling the midpoint, (2 × 3, 2 × 5), and then forgetting to take A off.
- (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.
- (d) 7.21 units — Method: the diagonal AC is the hypotenuse of the right-angled triangle ABC, whose shorter sides are AB and BC, so Pythagoras' theorem gives its length. Working: AB runs from (0, 0) to (6, 0), so AB = 6; BC runs from (6, 0) to (6, 4), so BC = 4. Then AC² = 6² + 4² = 36 + 16 = 52, so AC = √52 = 7.2111…, which is 7.21 correct to 2 decimal places. Answer: 7.21 units. The distractors: 10.00 units comes from adding the two sides, 6 + 4, instead of adding their squares and taking the root; 4.47 units comes from subtracting the squares, √(36 − 16), which is the form of Pythagoras used to find a shorter side rather than the hypotenuse; 26.00 units comes from halving 52 in place of taking its square root.
- (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.
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