Orienting Yourself: The Use of Coordinates — Detailed Notes
A coordinate system is a structured framework (like the grid lines on a map or graph paper) that lets us use numbers to describe the exact location of points. The idea has deep Indian roots — the grid-planned streets of the Sindhu-Sarasvatī cities, Baudhāyana’s N–S/E–W geometric constructions, Āryabhaṭa’s celestial coordinates, and Brahmagupta’s formalisation of zero and negative numbers (which make the four-quadrant plane possible). René Descartes (1637) finally showed every point in a plane can be fixed by two numbers — hence the ‘Cartesian’ plane.
1.3 The 2-D Cartesian coordinate system
Two number lines at right angles fix points in 2-D space. The horizontal line is the x-axis, the vertical line is the y-axis, and they meet at the origin O (0, 0). Distances to the right of O or up are positive; to the left or down are negative.
Coordinates (x, y): x (the abscissa) is the perpendicular distance of the point from the y-axis; y (the ordinate) is the perpendicular distance from the x-axis. Order matters: the x-coordinate is always written first.
• A point on the x-axis has the form (x, 0). • A point on the y-axis has the form (0, y). • The origin is (0, 0).
Quadrants
The axes divide the plane into four quadrants, numbered anticlockwise from the top-right:
| Quadrant | Sign of (x, y) | Example |
|---|
| I (top-right) | (+, +) | (3, 5) |
| II (top-left) | (−, +) | (−5, 3) |
| III (bottom-left) | (−, −) | (−4, −2) |
| IV (bottom-right) | (+, −) | (3, −5) |
Note: (x, y) and (y, x) are the same point only when x = y; otherwise they are different points (order matters).
1.4 Distance between two points
If a segment is parallel to an axis, its length is just the difference of the differing coordinates: distance between (x₁, y) and (x₂, y) is |x₂ − x₁|; between (x, y₁) and (x, y₂) is |y₂ − y₁|. For a slanting segment, drop perpendiculars to make a right triangle and use the Baudhāyana–Pythagoras theorem:
Distance = √[(x₂ − x₁)² + (y₂ − y₁)²]
It does not matter whether the differences are positive or negative — we are measuring shifts along the axes, and reflection in an axis preserves distances. Example: distance from A(3, 4) to D(7, 1) = √[(7−3)² + (1−4)²] = √(16 + 9) = √25 = 5 units.
Midpoint & useful results
• Midpoint M of the segment joining (x₁, y₁) and (x₂, y₂) is M = ((x₁ + x₂)/2, (y₁ + y₂)/2) — average the coordinates. • Collinearity (without plotting): three points are on one straight line if the ‘slope’ (rise/run) between each pair is equal — i.e. (y₂−y₁)/(x₂−x₁) = (y₃−y₂)/(x₃−x₂). • A point lies inside / on / outside a circle of centre O and radius r according as its distance from O is less than / equal to / greater than r.
Key Terms
coordinate system, Cartesian plane / xy-plane, x-axis, y-axis, origin, abscissa (x-coordinate), ordinate (y-coordinate), quadrant, distance formula, Baudhāyana–Pythagoras theorem, midpoint, section/trisection, collinear.
End-of-Chapter Exercises — Full Solutions
Complete worked solutions to all 16 end-of-chapter problems (including the starred HOTS problems). The in-text Exercise Sets 1.1 and 1.2 are figure-based room-plan activities to do on graph paper.
1What are the x-coordinate and y-coordinate of the point of intersection of the two axes?
The two axes intersect at the origin O. Both coordinates are 0, so the point is (0, 0).
2Point W has x-coordinate −5. Predict the coordinates of point H on the line through W parallel to the y-axis. Which quadrants can H lie in?
A line through W parallel to the y-axis is vertical, so every point on it has the same x-coordinate, i.e. x = −5. Thus H = (−5, y) for some value of y. Since x is negative, H lies to the left of the y-axis — it can be in Quadrant II (if y > 0) or Quadrant III (if y < 0), or on the x-axis (if y = 0).
3R(3, 0), A(0, −2), M(−5, −2), P(−5, 2) joined in order (RAMP). Predict (i) two perpendicular sides, (ii) a side parallel to an axis, (iii) two points that are mirror images in one axis.
(i) Side AM (from (0,−2) to (−5,−2)) is horizontal and side MP (from (−5,−2) to (−5,2)) is vertical, so AM ⊥ MP. (ii) AM is parallel to the x-axis (both ends have y = −2); also MP is parallel to the y-axis (both ends have x = −5). (iii) M(−5, −2) and P(−5, 2) are mirror images of each other in the x-axis (same x, opposite y). Plotting confirms all three.
4Plot Z(5, −6). Construct a right-angled triangle IZN and find the lengths of its three sides. (Answers may vary.)
One neat construction: take I(5, 0) and N(0, −6), with the right angle at Z. Then ZI (vertical) = |0 − (−6)| = 6, ZN (horizontal) = |5 − 0| = 5, and the hypotenuse IN = √[(5−0)² + (0−(−6))²] = √(25 + 36) = √61 ≈ 7.81 units. (Other valid constructions give different lengths.)
5What would a coordinate system be like without negative numbers? Could it locate all points of a 2-D plane?
Without negative numbers we could mark distances only to the right and up from O — i.e. only Quadrant I. Points to the left of or below the origin (Quadrants II, III, IV) could not be given coordinates. So no — such a system could not locate all points of the plane; negative numbers (Brahmagupta’s contribution) are essential.
6Are M(−3, −4), A(0, 0) and G(6, 8) on the same straight line? Suggest a method without plotting.
Method: compare the ‘slope’ (y-change ÷ x-change) between pairs of points; if equal, the points are collinear. Slope of AM = (−4−0)/(−3−0) = 4/3; slope of AG = (8−0)/(6−0) = 8/6 = 4/3. The slopes are equal, so yes, M, A and G are collinear.
7Use the same method for R(−5, −1), B(−2, −5) and C(4, −12).
Slope of RB = (−5−(−1))/(−2−(−5)) = (−4)/(3) = −4/3. Slope of BC = (−12−(−5))/(4−(−2)) = (−7)/(6) = −7/6. Since −4/3 ≠ −7/6, the slopes differ — so R, B and C are NOT collinear. Plotting confirms they form a triangle.
8Using the origin as one vertex, plot (i) a right-angled isosceles triangle, (ii) an isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV. (Sample answers.)
(i) O(0, 0), A(4, 0), B(0, 4): the right angle is at O and the two legs OA = OB = 4, so it is right-angled and isosceles. (ii) O(0, 0), P(−3, −4) in Quadrant III, Q(3, −4) in Quadrant IV: OP = √(9+16) = 5 and OQ = √(9+16) = 5, so OP = OQ — an isosceles triangle. (Other correct examples are possible.)
9For each row decide whether M is the midpoint of ST, and give the connection between the coordinates.
Using the midpoint formula M = ((xₛ+xₜ)/2, (yₛ+yₜ)/2): (−3,0),(0,0),(3,0) → mid = (0, 0) = M → Yes. (2,3),(3,4),(4,5) → mid = (3, 4) = M → Yes. (0,0),(0,5),(0,−10) → mid = (0, −5) ≠ (0, 5) → No. (−8,7),(0,−2),(6,−3) → mid = (−1, 2) ≠ (0, −2) → No. Connection: M is the midpoint when its coordinates are the averages of the corresponding coordinates of S and T.
10M(−7, 1) is the midpoint of A(3, −4) and B(x, y). Find B.
Using x = 2x_M − x_A and y = 2y_M − y_A: x = 2(−7) − 3 = −14 − 3 = −17; y = 2(1) − (−4) = 2 + 4 = 6. So B = (−17, 6). (Check: midpoint of (3,−4) and (−17,6) = (−7, 1). ✓)
11P, Q trisect AB (P closer to A, Q closer to B), with A(4, 7) and B(16, −2). Find P and Q.
P divides AB in ratio 1:2, so P = ((2·4 + 1·16)/3, (2·7 + 1·(−2))/3) = ((8+16)/3, (14−2)/3) = (24/3, 12/3) = (8, 4). Q divides AB in ratio 2:1, so Q = ((1·4 + 2·16)/3, (1·7 + 2·(−2))/3) = ((4+32)/3, (7−4)/3) = (36/3, 3/3) = (12, 1).
12(i) Show A(1, −8), B(−4, 7), C(−7, −4) lie on a circle K centred at O(0,0); find its radius. (ii) Do D(−5, 6) and E(0, 9) lie within, on, or outside K?
(i) OA = √(1² + 8²) = √65; OB = √(4² + 7²) = √65; OC = √(7² + 4²) = √65. All equal, so A, B, C lie on a circle of radius √65 ≈ 8.06 units. (ii) OD = √(25 + 36) = √61 < √65 → D is inside. OE = √(0 + 81) = 9 = √81 > √65 → E is outside.
13Midpoints of the sides of △ABC are D(5, 1), E(6, 5), F(0, 3). Find A, B, C.
Taking D, E, F as midpoints of the three sides, the vertices are given by adding two midpoints and subtracting the third: A = E + F − D = (6+0−5, 5+3−1) = (1, 7); B = F + D − E = (0+5−6, 3+1−5) = (−1, −1); C = D + E − F = (5+6−0, 1+5−3) = (11, 3). (Check: midpoint of A,B = (0,3)=F ✓, etc.)
14City grid, 10 streets each way; intersection of the a-th N–S street and b-th E–W street is called (a, b). Find how many intersections are (a) (4, 3) and (b) (3, 4).
Each label names one unique crossing — a single N–S street meeting a single E–W street. So (a) exactly 1 intersection is (4, 3), and (b) exactly 1 intersection is (3, 4). Note (4, 3) and (3, 4) are different crossings — order matters, just as with coordinates.
15Screen 800×600, origin at bottom-left. Circle A: centre (100, 150), r = 80. Circle B: centre (250, 230), r = 100. (i) Does any part lie outside the screen? (ii) Do the circles intersect?
(i) Circle A spans x: 20–180, y: 70–230 — all within 0–800 and 0–600, so it is fully on screen. Circle B spans x: 150–350, y: 130–330 — also fully on screen. No part of either circle is outside. (ii) Distance AB = √[(250−100)² + (230−150)²] = √(150² + 80²) = √(22500 + 6400) = √28900 = 170. Sum of radii = 80 + 100 = 180 and difference = 20. Since 20 < 170 < 180, the circles do intersect (at two points).
16Plot A(2, 1), B(−1, 2), C(−2, −1), D(1, −2). Is ABCD a square? Why? Find its area.
All four sides: AB = √[(2−(−1))²+(1−2)²] = √(9+1) = √10; BC = √[(−1−(−2))²+(2−(−1))²] = √(1+9) = √10; CD = √(9+1) = √10; DA = √(1+9) = √10 — all equal. Diagonals: AC = √[(2−(−2))²+(1−(−1))²] = √(16+4) = √20; BD = √[(−1−1)²+(2−(−2))²] = √(4+16) = √20 — equal. Four equal sides and equal diagonals → ABCD is a square. Area = side² = (√10)² = 10 square units.