Coordinate geometry: Questions and Answers

30 coordinate geometry questions solved step by step. Open a question for the full working, the marks for each step and exam practice.

• Let us examine the figure to understand the layout of the room. Notice that this only shows the map of the floor. Do you see why the position of the windows cannot be marked on this map?Answer: The map shows only the floor, which has two measurements (length and breadth). A window is set in a wall above the floor, so its height is a third measurement that a flat floor map cannot show.
• The figure shows Reiaan’s room with points OABC marking its corners. The x- and y-axes are marked in the figure. Point O is the origin.
  Referring to the figure, answer the following questions:Answer: (i) 8 ft from the left wall; 0 ft from the x-axis (the door lies on it) (ii) D₁ = (8, 0) (iii) 3.5 ft; yes, comfortable, and a wheelchair can pass easily (iv) the bathroom door is 2.5 ft wide, so it is narrower (by 1 ft)
• 1. What are the standard widths for a room door? Look around your home and in school.
  2. Are the doors in your school suitable for people in wheelchairs?Answer: 1. Room doors are usually about 2½ to 3 ft (75 to 90 cm) wide; bathroom doors about 2 to 2½ ft; main doors and classroom doors about 3 to 4 ft. 2. A door suits a wheelchair if its clear opening is at least about 3 ft (90 cm), with no step (or a ramp) and a handle within easy reach. Measure your school’s doors against this.
• So far, we have only considered points on the two coordinate axes. What can you say about the coordinates of points that are not on either axes?Answer: Neither coordinate is 0. The signs show which quadrant the point is in: (+, +) Quadrant I, (−, +) Quadrant II, (−, −) Quadrant III, (+, −) Quadrant IV.
• Copy the figure and mark S and Q in your diagram. Mark any point P in Quadrant I and any point R in Quadrant III, and write down their coordinates.Answer: S (3, −5): 3 right, 5 down. Q (−5, 3): 5 left, 3 up. Any point with both coordinates positive is in Quadrant I, e.g. P (4, 2); any point with both negative is in Quadrant III, e.g. R (−3, −4). (Answers for P and R will vary.)
• 1. What is the x-coordinate of a point on the y-axis?
  2. Is there a similar generalisation for a point on the x-axis?
  3. Does point Q (y, x) ever coincide with point P (x, y)? Justify your answer.
  4. If x ≠ y, then (x, y) ≠ (y, x); and (x, y) = (y, x) if and only if x = y. Is this claim true?Answer: 1. 0 2. Yes: every point on the x-axis has y-coordinate 0, so it is (x, 0). 3. Only when x = y, e.g. (3, 3). 4. Yes, the claim is true.
• Place Reiaan’s rectangular study table with three of its feet at the points (8, 9), (11, 9) and (11, 7).Answer: (i) (8, 7) (ii) Yes: it is clear of the bed, the wardrobe and the door’s swing, with open floor in front for a chair (iii) width 2 ft, length 3 ft; the height cannot be found from a floor plan
• If the bathroom door has a hinge at B₁ and opens into the bedroom, will it hit the wardrobe? Are there any changes you would suggest if the door is made wider?Answer: No. The door is 2.5 ft wide, so its edge swings on a circle of radius 2.5 ft about B₁ (0, 1.5); the wardrobe is 3 ft away (its side is at x = 3), leaving a 0.5 ft gap. A door 3 ft or wider would hit it, so then hinge it at the other end, make it open into the bathroom, use a sliding door, or move the wardrobe.
• Look at Reiaan’s bathroom.Answer: (i) O (0, 0), F (0, 9), R (−6, 9), P (−6, 0) (ii) a trapezium (SH ∥ RW): S (−6, 6), H (−3, 6), W (−2, 9), R (−6, 9) (iii) e.g. washbasin (−6, 0), (−4, 0), (−4, 3), (−6, 3); toilet (−6, 3.5), (−3, 3.5), (−3, 5.5), (−6, 5.5) (answers may vary)
• Other rooms in the house:Answer: (i) Corners P (−6, 0), A (12, 0), (12, −15), (−6, −15): the dining room lies below the x-axis (ii) centre (3, −7.5); table feet (0.5, −6), (5.5, −6), (5.5, −9), (0.5, −9)
• Look at triangle ADM in the figure. Triangle ADM is an acute angled triangle in the first quadrant. How do we find the lengths of its sides AD, DM and MA?Answer: Make a right-angled triangle on each side, with legs along the grid, and use the Baudhāyana–Pythagoras theorem: AD = √(4² + 3²) = 5 units, DM = √(2² + 5²) = √29 units, MA = √(6² + 2²) = √40 units.
• 1. In moving from A (3, 4) to D (7, 1), what distance has been covered along the x-axis? What about the distance along the y-axis?
  2. Can these distances help you find the distance AD?Answer: 1. Along the x-axis: 7 − 3 = 4 units (to the right); along the y-axis: 4 − 1 = 3 units (downwards). 2. Yes: they are the legs of a right-angled triangle with hypotenuse AD, so AD = √(4² + 3²) = 5 units.
• What if, x₁, x₂, y₁, y₂ take negative values? In the figure, triangle AMD is reflected in the y-axis. What are the coordinates of the images of points A, M, and D?Answer: A′ (−3, 4), M′ (−9, 6), D′ (−7, 1). Reflection in the y-axis changes the sign of the x-coordinate and keeps the y-coordinate. The distance formula still works with negative values: A′D′ = 5, D′M′ = √29, M′A′ = √40 units, the same as before.
• 1. What has remained the same and what has changed with this reflection?
  2. Would these observations be the same if ΔADM is reflected in the x-axis (instead of the y-axis)?Answer: 1. Same: side lengths (5, √29, √40), angles, shape, size and area (13 sq units), and every y-coordinate. Changed: every x-coordinate changes sign, the triangle moves from Quadrant I to Quadrant II, and it is flipped (mirror image). 2. Yes, the same kind of observations: lengths, angles and area stay the same, but now the x-coordinates stay, the y-coordinates change sign, and the triangle moves to Quadrant IV.
• What are the x-coordinate and y-coordinate of the point of intersection of the two axes?Answer: The axes meet at the origin O: x-coordinate 0 and y-coordinate 0, so O = (0, 0).
• Point W has x-coordinate equal to −5. Can you predict the coordinates of point H which is on the line through W parallel to the y-axis? Which quadrants can H lie in?Answer: H = (−5, k) for some number k: its x-coordinate must be −5, its y-coordinate can be anything. H lies in Quadrant II (if k > 0) or Quadrant III (if k < 0); if k = 0, H = (−5, 0) is on the x-axis.
• Consider the points R (3, 0), A (0, −2), M (−5, −2) and P (−5, 2). If they are joined in the same order, predict:Answer: (i) AM ⊥ MP (ii) AM ∥ x-axis (also MP ∥ y-axis) (iii) M (−5, −2) and P (−5, 2), mirror images in the x-axis. The plot confirms all three.
• Plot point Z (5, −6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides.
  (Comment: Answers may differ from person to person.)Answer: One answer: N (5, −2) and I (2, −2), with the right angle at N. ZN = 4, NI = 3, IZ = √(3² + 4²) = 5 units. (Other choices are also correct.)
• What would a system of coordinates be like if we did not have negative numbers? Would this system allow us to locate all the points on a 2-D plane?Answer: Only the first quadrant (with the positive halves of the axes) could be described: points to the right of and above the origin. No, it could not locate all points: nothing to the left of the y-axis or below the x-axis could be named.
• Are the points M (−3, −4), A (0, 0) and G (6, 8) on the same straight line? Suggest a method to check this without plotting and joining the points.Answer: Yes. Method: find the three distances. MA = 5, AG = 10, MG = 15, and MA + AG = MG, so A lies on the segment MG and the three points are on one straight line.
• Use your method (the one used for M (−3, −4), A (0, 0) and G (6, 8)) to check if the points R (−5, −1), B (−2, −5) and C (4, −12) are on the same straight line. Now plot both sets of points and check your answers.Answer: No. RB = 5, BC = √85 ≈ 9.220, RC = √202 ≈ 14.213, and RB + BC ≈ 14.220 ≠ RC. They are not on one line, although on a plot they look almost straight; M, A, G (the earlier set) plot exactly on one line.
• Using the origin as one vertex, plot the vertices of:Answer: (i) e.g. O (0, 0), A (4, 0), B (0, 4): OA = OB = 4 and the angle at O is 90° (ii) e.g. O (0, 0), P (−3, −4), Q (3, −4): OP = OQ = 5 (answers may vary)
• The following table shows the coordinates of points S, M and T. In each case, state whether M is the midpoint of segment ST. Justify your answer.Answer: Row 1: Yes (SM = MT = 3, ST = 6) · Row 2: Yes (SM = MT = √2, ST = 2√2) · Row 3: No (SM = 5, MT = 15) · Row 4: No (SM = √145, MT = √37). Connection: the midpoint’s coordinates are the averages, M = (x_S + x_T2, y_S + y_T2).
• Use the connection you found to find the coordinates of B given that M (−7, 1) is the midpoint of A (3, −4) and B (x, y).Answer: B = (−17, 6), since 3 + x2 = −7 gives x = −17 and −4 + y2 = 1 gives y = 6.
• Let P, Q be points of trisection of AB, with P closer to A, and Q closer to B. Using your knowledge of how to find the coordinates of the midpoint of a segment, how would you find the coordinates of P and Q? Do this for the case when the points are A (4, 7) and B (16, −2).Answer: Since AP = PQ = QB, P is the midpoint of AQ and Q is the midpoint of PB; solving these gives P = (2x₁ + x₂3, 2y₁ + y₂3) and Q = (x₁ + 2x₂3, y₁ + 2y₂3). For A (4, 7), B (16, −2): P (8, 4) and Q (12, 1).
• (i) Given the points A (1, −8), B (−4, 7) and C (−7, −4), show that they lie on a circle K whose center is the origin O (0, 0). What is the radius of circle K?
  (ii) Given the points D (−5, 6) and E (0, 9), check whether D and E lie within the circle, on the circle, or outside the circle K.Answer: (i) OA = OB = OC = √65, so A, B, C lie on a circle with centre O; radius = √65 units (≈ 8.06) (ii) OD = √61 < √65, so D is inside; OE = 9 = √81 > √65, so E is outside
• The midpoints of the sides of triangle ABC are the points D, E, and F. Given that the coordinates of D, E, and F are (5, 1), (6, 5), and (0, 3), respectively, find the coordinates of A, B and C.Answer: Taking D, E, F as the midpoints of BC, CA, AB: A (1, 7), B (−1, −1), C (11, 3).
• A city has two main roads which cross each other at the centre of the city. These two roads are along the North–South (N–S) direction and East–West (E–W) direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are 10 streets in each direction.Answer: (i) Draw 10 vertical and 10 horizontal lines, 1 cm apart (a 9 cm × 9 cm grid) (ii) (a) only one intersection is (4, 3) (b) only one intersection is (3, 4), and it is a different one
• A computer graphics program displays images on a rectangular screen whose coordinate system has the origin at the bottom-left corner. The screen is 800 pixels wide and 600 pixels high. A circular icon of radius 80 pixels is drawn with its centre at the point A (100, 150). Another circular icon of radius 100 pixels is drawn with its centre at the point B (250, 230). Determine:Answer: (i) No. Circle A spans x 20 to 180, y 70 to 230; circle B spans x 150 to 350, y 130 to 330; both lie inside 0–800 × 0–600 (ii) Yes. AB = √(150² + 80²) = 170 < 80 + 100 = 180, so the circles overlap and cross at two points
• Plot the points A (2, 1), B (−1, 2), C (−2, −1), and D (1, −2) in the coordinate plane. Is ABCD a square? Can you explain why? What is the area of this square?Answer: Yes. All four sides are √10 and both diagonals are √20, so ABCD is a square. Area = (√10)² = 10 square units.

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