Printable · GCSE Foundation · ages 14-16
Coordinates in all four quadrants worksheet — GCSE Foundation
Fifteen questions on "coordinates in all four quadrants" — DfE statement A8. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Answer key: Coordinates in all four quadrants worksheet — GCSE Foundation
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- (b) 3 — −6 + 9 = 3. A candidate who ignores the negative sign on the x-coordinate and adds the two positive values gets 6 + 9 = 15. A candidate who treats the y-coordinate as negative too gets −6 + (−9) = −15. A candidate who works out 9 − 6 correctly as 3 but then writes the answer with the wrong sign gets −3.
- (a) 5.8 — The horizontal distance is 3 and the vertical distance is 5, so using Pythagoras' theorem the distance is √(3² + 5²) = √34 = 5.8 (1 d.p.). A candidate who adds the two differences instead of using Pythagoras gets 3 + 5 = 8.0. A candidate who works out 3² + 5² = 34 but forgets to take the square root gets 34.0. A candidate who subtracts the squares instead of adding them gets √(5² − 3²) = √16 = 4.0.
- (c) 4 — Method: the distance of a point from the x-axis is measured vertically, so it is the size of the y-coordinate taken without its sign. Working: the point (−6, 4) has y-coordinate 4, so moving straight down to the x-axis covers 4 units, and a distance is written as a positive number. Answer: 4. The distractors: 6 comes from using the x-coordinate, which measures the distance from the y-axis rather than from the x-axis; −6 comes from that same mistake with the minus sign left in place, although a distance is never negative; 10 comes from adding the two distances, 6 and 4, as though the question asked how far the point is from both axes together.
- (d) the third quadrant — Method: only the signs of the coordinates matter, so two coordinates equal in size still have to be read separately, one for each axis. Working: both coordinates of (−2, −2) are negative, so B lies to the left of the y-axis and below the x-axis; counting anticlockwise from the region where both coordinates are positive, left and below is the third region. Answer: the third quadrant. The distractors: 'the first quadrant' comes from ignoring both minus signs and treating the point as (2, 2); 'the second quadrant' comes from applying the minus sign to the x-coordinate only, as though the point were (−2, 2); 'the fourth quadrant' comes from applying it to the y-coordinate only, as though the point were (2, −2).
- (b) (−5, −2) — A rotation of 180° about the origin reverses the sign of both coordinates, so Q = (−5, −2). A candidate who reverses the sign of only the y-coordinate, as if reflecting in the x-axis, gets (5, −2). A candidate who reverses the sign of only the x-coordinate, as if reflecting in the y-axis, gets (−5, 2). A candidate who swaps the coordinates instead of reversing their signs gets (2, 5).
- (a) (2, 7) — Midpoint = ((x1+x2)/2, (y1+y2)/2) = ((−3+7)/2, (5+9)/2) = (4/2, 14/2) = (2, 7). (4, 14) comes from adding the coordinates correctly but forgetting to divide by 2. (2, 9) comes from correctly averaging the x-coordinates but simply copying the y-coordinate of the second point instead of averaging the y-coordinates. (5, 2) comes from subtracting the coordinates instead of adding them before halving: ((7−(−3))/2, (9−5)/2) = (5, 2).
- (d) the y-axis, and 12 — Method: coordinates are written (x, y), so the first number is the distance across and the second the distance up; a point whose first coordinate is 0 has not moved across from the origin and therefore lies on the vertical axis. Working: in (0, 12) the first number is 0, so P is on the y-axis, and the second number, 12, is the y-coordinate of P. Answer: the y-axis, and 12. The distractors: 'the x-axis, and 12' comes from mixing up which axis the condition 'the first coordinate is 0' describes; 'the y-axis, and 0' comes from placing P correctly but reading the pair the wrong way round, so that the first number is quoted as the y-coordinate; 'the x-axis, and 0' comes from making both of those mistakes at once.
- (a) (−6, −3) — Reflecting in the y-axis changes the sign of the x-coordinate and keeps the y-coordinate the same: (−6, −3). (6, 3) comes from reflecting in the x-axis instead, which changes the sign of the y-coordinate. (−6, 3) comes from reflecting in both axes. (6, −3) comes from not applying the reflection at all.
- (b) (−1, 6) — '6 units above the x-axis' gives a y-coordinate of 6, and the x-coordinate is given as −1, so F = (−1, 6). (6, −1) comes from swapping the x- and y-coordinates. (−1, −6) comes from treating 'above' as a negative direction instead of positive. (1, 6) comes from dropping the negative sign on the given x-coordinate.
- (d) (4, 8) — The other endpoint is found from 2 × midpoint − known endpoint: x = 2 × (−1) − (−6) = −2 + 6 = 4, y = 2 × 5 − 2 = 10 − 2 = 8, giving (4, 8). (−3.5, 3.5) comes from averaging the given endpoint and the midpoint as if they were the two endpoints of a segment, instead of working backwards from the midpoint. (5, 3) comes from working out (−1 − (−6), 5 − 2) instead of doubling the midpoint before subtracting. (4, 5) comes from correctly finding the x-coordinate but copying the midpoint's y-coordinate of 5 instead of doubling it.
- (d) (−1, 4) — In parallelogram ABCD the side DC is parallel and equal to the side AB, so D = C − AB. The vector from A to B is (2 − (−3), 1 − 1) = (5, 0), so D = (4 − 5, 4 − 0) = (−1, 4). A candidate who adds this vector to C instead of subtracting it gets (4 + 5, 4 + 0) = (9, 4). A candidate who subtracts A's coordinates from C's rather than the vector AB, and drops the minus sign on −3 while doing so, works out (4 − 3, 4 − 1) and gets (1, 3). A candidate who makes only the y-part of that slip, working out 4 − 1 instead of 4 − 0, gets (−1, 3).
- (d) 10 — The right angle is at (1, 1). The vertical side has length 5 − 1 = 4 and the horizontal side has length 6 − 1 = 5, so the area is (4 × 5) ÷ 2 = 20 ÷ 2 = 10. A candidate who forgets to halve the product of the two sides gets 4 × 5 = 20. A candidate who forgets to subtract the shared vertex's coordinate and uses the raw coordinates 6 and 5 as the side lengths gets (6 × 5) ÷ 2 = 30 ÷ 2 = 15. A candidate who uses only one side length as the area gets 5.
- (b) (−4, 3) — Method: in a rectangle whose sides are parallel to the axes only two different x-coordinates and two different y-coordinates appear, and each of them is shared by a pair of vertices, so the missing vertex takes the x-coordinate and the y-coordinate that so far appear only once. Working: the x-coordinates given are −4, 2 and 2, so 2 is already used twice and −4 is used once; the y-coordinates given are −1, −1 and 3, so −1 is already used twice and 3 is used once; the fourth vertex therefore has x = −4 and y = 3. Answer: (−4, 3). The distractors: (3, −4) comes from picking the two unpaired coordinates correctly and then writing them in the wrong order; (−4, −5) comes from matching the 4-unit vertical side but measuring it downwards from (−4, −1) instead of upwards; (8, 3) comes from carrying on round the shape with the horizontal step used earlier, adding 6 to the x-coordinate of (2, 3) instead of closing the rectangle.
- (c) (−4, 7) — Left of the origin means negative x, and up means positive y, so the point is (−4, 7). (4, 7) comes from forgetting that 'left' means the x-coordinate is negative. (−4, −7) comes from treating 'up' as a negative direction instead of positive. (7, −4) comes from swapping the x- and y-coordinates.
- (a) 8 — Both points share the x-coordinate, so the distance between them is the difference between the y-coordinates: 3 − (−5) = 8. A candidate who mistakenly uses the equal x-coordinates instead of the y-coordinates gets 4 − 4 = 0. A candidate who adds the y-coordinates instead of subtracting them gets 3 + (−5) = −2. A candidate who reads off only point B's y-coordinate as the distance gets 3.
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