1. Chapter at a Glance
How do you tell a friend exactly where you are sitting in a cinema hall? "Row F, seat 12." Two pieces of information fix one position. This simple idea, developed into mathematics by René Descartes in the 17th century, gives us the Cartesian plane — a grid on which every point is named by an ordered pair of numbers. Once points have numbers, geometry becomes algebra: we can calculate distances, midpoints and check whether a shape is a square or a right triangle, without ever measuring with a ruler.
- The coordinate axes, origin, quadrants and ordered pairs.
- Plotting and reading points; points on the axes.
- Distance between two points (using the Baudhāyana–Pythagoras theorem).
- Midpoint of a line segment.
- Using coordinates to prove properties of triangles and quadrilaterals.
- Collinearity and simple reflections.
2. The Cartesian Plane
Draw a horizontal number line (the x-axis) and a vertical number line (the y-axis) crossing at their zeros. The crossing point is the origin O(0, 0). The axes divide the plane into four quadrants, numbered anticlockwise starting from the top right.
- A point is written as an ordered pair (x, y). x is the abscissa (horizontal distance from the y-axis); y is the ordinate (vertical distance from the x-axis).
- Order matters: (3, 2) and (2, 3) are different points.
- Any point on the x-axis has the form (a, 0); on the y-axis, (0, b).
| Quadrant | Sign of x | Sign of y | Example |
|---|---|---|---|
| I | + | + | (4, 1) |
| II | − | + | (−2, 5) |
| III | − | − | (−3, −3) |
| IV | + | − | (6, −1) |
| On an axis | x = 0 or y = 0 | (0, −4), (7, 0) | |
3. Distance Between Two Points
Horizontal and vertical distances first. Points with the same y-coordinate lie on a horizontal line; the distance between them is |x₂ − x₁|. Points with the same x-coordinate lie on a vertical line; the distance is |y₂ − y₁|.
Slanting distance. For A(x₁, y₁) and B(x₂, y₂), draw a horizontal line through A and a vertical line through B. They meet at C(x₂, y₁), forming a right triangle with legs |x₂ − x₁| and |y₂ − y₁|. By the Baudhāyana–Pythagoras theorem:
AB = √[(x₂ − x₁)² + (y₂ − y₁)²]
Distance of P(x, y) from the origin: OP = √(x² + y²).
Find the distance between A(−2, 3) and B(4, −5).
Solution: x₂ − x₁ = 6, y₂ − y₁ = −8. AB = √(36 + 64) = √100 = 10 units.
The distance between P(x, 2) and Q(3, −4) is 10. Find x.
Solution: (3 − x)² + 36 = 100 → (3 − x)² = 64 → 3 − x = ±8 → x = −5 or x = 11. Two points fit — one on each side.
Find the point on the x-axis equidistant from A(2, 5) and B(−2, 9).
Solution: Let P = (a, 0). PA² = PB² → (a − 2)² + 25 = (a + 2)² + 81 → −4a + 4 + 25 = 4a + 4 + 81 → −8a = 56 → a = −7. P(−7, 0).
4. Midpoint of a Segment
The midpoint M of AB is halfway in the x-direction and halfway in the y-direction, so its coordinates are the averages:
M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )
Find the midpoint of A(−3, 8) and B(5, −2).
Solution: M = ((−3 + 5)/2, (8 − 2)/2) = (1, 3).
M(2, −1) is the midpoint of AB, and A = (6, 3). Find B.
Solution: (6 + x)/2 = 2 → x = −2; (3 + y)/2 = −1 → y = −5. B(−2, −5).
Centroid (enrichment): the medians of a triangle meet at the centroid G = ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3).
5. Proving Shapes with Coordinates
Coordinates let us prove what a figure is, rather than just guess from a drawing.
| To show the figure is a… | Check |
|---|---|
| Isosceles triangle | Two sides equal |
| Equilateral triangle | All three sides equal |
| Right triangle | Largest side² = sum of the squares of the other two |
| Parallelogram | Diagonals have the same midpoint (bisect each other) |
| Rectangle | Parallelogram + equal diagonals |
| Rhombus | All four sides equal |
| Square | All sides equal + equal diagonals |
Show that A(1, 1), B(4, 4) and C(7, 1) form a right isosceles triangle.
Solution: AB² = 9 + 9 = 18; BC² = 9 + 9 = 18; AC² = 36 + 0 = 36. AB = BC (isosceles) and AB² + BC² = 36 = AC² (right angle at B). ✓
A(1, 2), B(5, 3), C(6, 7) are three vertices of parallelogram ABCD. Find D.
Solution: Diagonals AC and BD share a midpoint. Midpoint of AC = (3.5, 4.5). So (5 + x)/2 = 3.5 → x = 2; (3 + y)/2 = 4.5 → y = 6. D(2, 6).
Collinear points: three points A, B, C lie on one line if the two shorter distances add up to the longest: AB + BC = AC.
6. Reflections in the Axes
| Reflection of P(x, y) in… | Image |
|---|---|
| x-axis | (x, −y) |
| y-axis | (−x, y) |
| Origin | (−x, −y) |
7. Solved Problems (Board Pattern)
Find the perimeter of the triangle with vertices O(0, 0), A(6, 0) and B(0, 8).
Solution: OA = 6, OB = 8, AB = √(36 + 64) = 10. Perimeter = 24 units.
Check whether A(0, 0), B(3, 1), C(2, 4), D(−1, 3) form a square.
Solution: AB² = 10, BC² = 1 + 9 = 10, CD² = 9 + 1 = 10, DA² = 1 + 9 = 10 → all sides equal. Diagonals: AC² = 4 + 16 = 20, BD² = 16 + 4 = 20 → equal. So ABCD is a square of side √10.
Are A(1, 2), B(3, 6), C(4, 8) collinear?
Solution: AB = √20 = 2√5, BC = √5, AC = √45 = 3√5. AB + BC = 3√5 = AC, so yes.
8. Master Formula Table
| Result | Formula |
|---|---|
| Distance AB | √[(x₂ − x₁)² + (y₂ − y₁)²] |
| Distance from origin | √(x² + y²) |
| Horizontal distance | |x₂ − x₁| |
| Vertical distance | |y₂ − y₁| |
| Midpoint | ((x₁ + x₂)/2, (y₁ + y₂)/2) |
| Centroid | ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3) |
| Point on x-axis / y-axis | (a, 0) / (0, b) |
| Reflections | x-axis (x, −y); y-axis (−x, y); origin (−x, −y) |
9. Master Summary Table
| Topic | Key idea | Exam focus |
|---|---|---|
| Ordered pair | (abscissa, ordinate); order matters | Identify quadrant |
| Quadrants | Signs (+,+), (−,+), (−,−), (+,−) | Sign-based MCQs |
| Distance | Pythagoras on a grid | Find unknown coordinate |
| Midpoint | Average of coordinates | Missing endpoint, fourth vertex |
| Shape proofs | Compare sides and diagonals | Square, rhombus, right triangle |
| Collinearity | AB + BC = AC | Three-point checks |
Chart — distances of points from the origin:
10. Mind Map
- Axes, origin
- 4 quadrants
- √(Δx² + Δy²)
- Unknown coordinate
- Average
- Fourth vertex
- Triangles, quadrilaterals
- Collinearity
11. Tips and Tricks
- Compare squared distances to avoid square roots when checking equality or right angles.
- Spot Pythagorean triples: differences (3, 4), (6, 8), (5, 12) give distances 5, 10, 13.
- For a parallelogram's fourth vertex: D = A + C − B (add coordinates of opposite vertices, subtract the third).
- Swapping x and y when plotting.
- Sign errors: (−3 − 5)² = 64, not 4.
- Equal sides alone do not make a square — a rhombus also has equal sides. Check diagonals too.
12. Important One-liners
- The origin is (0, 0); every point on the y-axis has x = 0.
- A point (−, +) lies in Quadrant II.
- Distance from the origin to (x, y) is √(x² + y²).
- The midpoint's coordinates are the averages of the endpoints' coordinates.
- The diagonals of a parallelogram bisect each other — they have the same midpoint.
- Coordinate geometry was developed by René Descartes (and independently by Pierre de Fermat).
13. Exam-pattern Questions with Detailed Answers
In which quadrant does (−4, −7) lie?
- I
- II
- III
- IV
Show answer
Both coordinates are negative → Quadrant III.
Answer: (C)
Find the distance of (−6, 8) from the origin.
Show answer
√(36 + 64) = 10.
Answer: 10 units
Find the midpoint of (2, −6) and (8, 4).
Show answer
((2 + 8)/2, (−6 + 4)/2) = (5, −1).
Answer: (5, −1)
Find the distance between (a, b) and (−a, −b).
Show answer
√[(2a)² + (2b)²] = 2√(a² + b²).
Answer: 2√(a² + b²)
Find y if the distance between P(2, −3) and Q(10, y) is 10.
Show answer
64 + (y + 3)² = 100 → (y + 3)² = 36 → y + 3 = ±6 → y = 3 or y = −9.
Answer: y = 3 or −9
Find the point on the y-axis equidistant from A(6, 5) and B(−4, 3).
Show answer
Let P = (0, k). 36 + (5 − k)² = 16 + (3 − k)² → 36 + 25 − 10k = 16 + 9 − 6k → 36 = 4k → k = 9.
Answer: (0, 9)
The midpoint of the segment joining A(2a, 4) and B(−2, 3b) is M(1, 2a + 1). Find a and b.
Show answer
(2a − 2)/2 = 1 → a = 2. (4 + 3b)/2 = 2a + 1 = 5 → 3b = 6 → b = 2.
Answer: a = 2, b = 2
Show that (5, −2), (6, 4) and (7, −2) are the vertices of an isosceles triangle.
Show answer
Sides² : 1 + 36 = 37, 1 + 36 = 37, 4 + 0 = 4. Two sides equal (√37), so isosceles.
Answer: Two sides = √37 → isosceles
Show that A(−1, −2), B(1, 0), C(−1, 2), D(−3, 0) are the vertices of a square.
Show answer
Each side² = 4 + 4 = 8 (AB, BC, CD, DA). Diagonals: AC² = 0 + 16 = 16, BD² = 16 + 0 = 16. Equal sides and equal diagonals → square.
Answer: It is a square of side 2√2
A(1, 3), B(−1, 2) and C(2, 5) are three consecutive vertices of parallelogram ABCD. Find D and verify that AB = CD.
Show answer
Diagonals AC and BD bisect each other. Midpoint of AC = (1.5, 4). So (−1 + x)/2 = 1.5 → x = 4 and (2 + y)/2 = 4 → y = 6. D = (4, 6). Check: AB² = 4 + 1 = 5, CD² = 4 + 1 = 5 ✓.
Answer: D(4, 6)
Find the ratio in which the y-axis divides the segment joining P(−3, 5) and Q(6, −1), and the point of division.
Show answer
The x-coordinate goes from −3 to 6 (change 9) and reaches 0 after 3 units. So the y-axis cuts PQ in ratio 3 : 6 = 1 : 2. The y-coordinate then is 5 + (1/3)(−6) = 3. Point: (0, 3).
Answer: 1 : 2 at (0, 3)
Prove using coordinates that the midpoint of the hypotenuse of a right triangle is equidistant from all three vertices. (Take O(0, 0), A(2a, 0), B(0, 2b).)
Show answer
Midpoint of AB: M(a, b). MO = √(a² + b²). MA = √[(2a − a)² + b²] = √(a² + b²). MB = √[a² + (2b − b)²] = √(a² + b²). All equal. ∎
Answer: MO = MA = MB = √(a² + b²)
14. Quick Revision Summary
- Point = (x, y); quadrant signs I(+,+), II(−,+), III(−,−), IV(+,−).
- Distance = √[(x₂ − x₁)² + (y₂ − y₁)²]; from origin √(x² + y²).
- Midpoint = average of coordinates; parallelogram diagonals share a midpoint.
- Prove shapes by comparing sides and diagonals; collinear if AB + BC = AC.
- Reflections: change the sign of one or both coordinates.