1. Chapter at a Glance
A town planner marks a bus stop, a school and a hospital on a grid map. Is the school exactly halfway between the other two? Where should a new road junction be placed so that it divides a road in the ratio 2 : 3? Coordinate geometry answers such questions with simple formulas. In this chapter you will use the distance formula and the section formula (including the midpoint formula) to solve geometry problems algebraically. The chapter carries about 6 marks in the CBSE board paper.
- Distance between two points; distance from the origin.
- Applying distance to check collinearity and identify triangles/quadrilaterals.
- Section formula for internal division in the ratio m₁ : m₂.
- Midpoint formula and points of trisection.
- Finding the ratio in which a point or an axis divides a segment.
2. Distance Formula
Distance between P(x₁, y₁) and Q(x₂, y₂):
Distance of P(x, y) from the origin: OP = √(x² + y²).
Find the distance between (−5, 7) and (−1, 3).
Solution: √(16 + 16) = √32 = 4√2.
Do A(1, 5), B(2, 3), C(−2, −11) lie on a line?
Solution: AB = √5, BC = √(16 + 196) = √212, AC = √(9 + 256) = √265. √5 + √212 ≈ 2.24 + 14.56 = 16.80 ≠ √265 ≈ 16.28. Not collinear.
Find a relation between x and y such that P(x, y) is equidistant from A(7, 1) and B(3, 5).
Solution: (x − 7)² + (y − 1)² = (x − 3)² + (y − 5)² → −14x − 2y + 50 = −6x − 10y + 34 → x − y = 2.
3. Section Formula
Point P divides AB internally in the ratio m₁ : m₂ if AP : PB = m₁ : m₂.
Memory aid: "cross-multiply" — m₁ goes with the far point B(x₂, y₂), m₂ with A(x₁, y₁).
Why it works: drop perpendiculars to make two similar right triangles; their sides are in ratio m₁ : m₂, giving (x − x₁)/(x₂ − x) = m₁/m₂, which solves to the formula.
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Find the point dividing the join of (4, −3) and (8, 5) in the ratio 3 : 1 internally.
Solution: x = (3 × 8 + 1 × 4)/4 = 7; y = (3 × 5 + 1 × (−3))/4 = 3. (7, 3).
Find the points of trisection of the segment joining A(2, −2) and B(−7, 4).
Solution: Ratio 1 : 2: ((−7 + 4)/3, (4 − 4)/3) = (−1, 0). Ratio 2 : 1: ((−14 + 2)/3, (8 − 2)/3) = (−4, 2).
4. Finding the Ratio
Take the ratio as k : 1. Write the coordinates of the dividing point using the section formula and match with the given condition.
In what ratio does (−4, 6) divide the join of A(−6, 10) and B(3, −8)?
Solution: x: (3k − 6)/(k + 1) = −4 → 3k − 6 = −4k − 4 → k = 2/7. Ratio 2 : 7. (Check y: (−16/7 + 10)/(9/7) = (54/7)/(9/7) = 6 ✓.)
Find the ratio in which the y-axis divides the join of (5, −6) and (−1, −4), and the point of division.
Solution: On the y-axis x = 0: (−k + 5)/(k + 1) = 0 → k = 5. Ratio 5 : 1. y = (−20 − 6)/6 = −13/3. Point (0, −13/3).
5. Solved Problems (Board Pattern)
(1, 2), (4, y), (x, 6), (3, 5) are vertices of a parallelogram in order. Find x and y.
Solution: Diagonals bisect each other: midpoint of (1, 2)&(x, 6) = midpoint of (4, y)&(3, 5). (1 + x)/2 = 7/2 → x = 6; 8/2 = (y + 5)/2 → y = 3.
Show that (3, 0), (4, 5), (−1, 4), (−2, −1) form a rhombus.
Solution: All sides = √26 (e.g. √(1 + 25)). Diagonals: √(16 + 16) = 4√2 and √(36 + 36) = 6√2 — unequal. Equal sides, unequal diagonals → rhombus (not a square).
Find the points dividing the join of A(−2, 2) and B(2, 8) into four equal parts.
Solution: Midpoint M = (0, 5). Midpoint of AM = (−1, 7/2); midpoint of MB = (1, 13/2). Points: (−1, 3.5), (0, 5), (1, 6.5).
6. Master Formula Table
| Result | Formula |
|---|---|
| Distance | √[(x₂ − x₁)² + (y₂ − y₁)²] |
| From origin | √(x² + y²) |
| Section (m₁ : m₂) | ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)) |
| Midpoint | ((x₁ + x₂)/2, (y₁ + y₂)/2) |
| Ratio k : 1 | ((kx₂ + x₁)/(k + 1), (ky₂ + y₁)/(k + 1)) |
| Centroid (enrichment) | ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3) |
7. Master Summary Table
| Question type | Tool | Check |
|---|---|---|
| Length / perimeter | Distance formula | Simplify surds |
| Type of triangle | Compare sides; Pythagoras | Isosceles, right, equilateral |
| Type of quadrilateral | Sides + diagonals | Square vs rhombus vs rectangle |
| Dividing point | Section formula | Correct pairing of m₁, m₂ |
| Find ratio | k : 1 method | Verify with other coordinate |
| Missing vertex | Midpoints of diagonals | Order of vertices |
8. Mind Map
- √(Δx² + Δy²)
- Collinearity
- m₁ : m₂
- Cross pairing
- Average
- Trisection
- Triangles
- Parallelogram, rhombus
9. Tips and Tricks
- Use k : 1 for unknown ratios — only one unknown to solve.
- A point on the x-axis has y = 0; on the y-axis x = 0 — use this to find ratios.
- Compare squared distances to avoid square roots.
- Pairing m₁ with (x₁, y₁) instead of (x₂, y₂).
- Sign errors with negative coordinates: (−1 − (−5)) = 4.
- Concluding "square" from equal sides alone.
10. Important One-liners
- The distance formula comes from the Baudhāyana–Pythagoras theorem.
- The midpoint divides a segment in the ratio 1 : 1.
- Points of trisection divide a segment in 1 : 2 and 2 : 1.
- The x-axis divides a segment where the y-coordinate becomes 0.
- Diagonals of a parallelogram have the same midpoint.
11. Exam-pattern Questions with Detailed Answers
The distance of (−6, 8) from the origin is:
- 6
- 8
- 10
- 14
Show answer
√(36 + 64) = 10.
Answer: (C)
Find the midpoint of (−1, 7) and (5, −3).
Show answer
(2, 2).
Answer: (2, 2)
Find the distance between (a, b) and (−a, −b).
Show answer
√(4a² + 4b²) = 2√(a² + b²).
Answer: 2√(a² + b²)
Find y if the distance between P(2, −3) and Q(10, y) is 10 units.
Show answer
64 + (y + 3)² = 100 → y + 3 = ±6 → y = 3 or −9.
Answer: 3 or −9
Find the coordinates of the point dividing the join of (−1, 7) and (4, −3) in the ratio 2 : 3.
Show answer
x = (8 − 3)/5 = 1; y = (−6 + 21)/5 = 3. (1, 3).
Answer: (1, 3)
Find the ratio in which the x-axis divides the segment joining (1, −5) and (−4, 5).
Show answer
(5k − 5)/(k + 1) = 0 → k = 1. Ratio 1 : 1; point (−3/2, 0).
Answer: 1 : 1 at (−1.5, 0)
Find the point on the x-axis equidistant from (2, −5) and (−2, 9).
Show answer
(x − 2)² + 25 = (x + 2)² + 81 → −8x = 56 → x = −7. (−7, 0).
Answer: (−7, 0)
Show that (1, 7), (4, 2), (−1, −1), (−4, 4) are vertices of a square.
Show answer
Sides: √(9 + 25) = √34 each (four sides). Diagonals: (1, 7)–(−1, −1): √(4 + 64) = √68; (4, 2)–(−4, 4): √(64 + 4) = √68. Equal sides and equal diagonals → square.
Answer: It is a square
If A(−2, −1), B(a, 0), C(4, b) and D(1, 2) are vertices of a parallelogram in order, find a and b.
Show answer
Midpoint AC = midpoint BD: (−2 + 4)/2 = (a + 1)/2 → a = 1; (−1 + b)/2 = (0 + 2)/2 → b = 3.
Answer: a = 1, b = 3
P(−1, 2) and Q(1, 0) are the points of trisection of AB, in the order A, P, Q, B. Find A and B.
Show answer
P is the midpoint of AQ: A = 2P − Q = (−3, 4). Q is the midpoint of PB: B = 2Q − P = (3, −2).
Answer: A(−3, 4), B(3, −2)
Find the ratio in which the line segment joining A(1, −5) and B(−4, 5) is divided by the point P(x, 2) — and find x.
Show answer
y: (5k − 5)/(k + 1) = 2 → 5k − 5 = 2k + 2 → k = 7/3. Ratio 7 : 3. x = (7(−4) + 3(1))/10 = −25/10 = −5/2.
Answer: 7 : 3; x = −5/2
Show that the points A(1, −2), B(3, 6), C(5, 10), D(3, 2) are the vertices of a parallelogram but not a rectangle.
Show answer
Midpoint AC = (3, 4); midpoint BD = (3, 4) → diagonals bisect each other → parallelogram. AC = √(16 + 144) = √160; BD = √(0 + 16) = 4. Diagonals unequal → not a rectangle.
Answer: Parallelogram, not a rectangle
12. Quick Revision Summary
- Distance = √[(x₂ − x₁)² + (y₂ − y₁)²].
- Section formula: ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)).
- Midpoint = average; trisection = ratios 1 : 2 and 2 : 1.
- Unknown ratio → take k : 1; axis → set x or y to 0.
- Identify shapes by comparing sides and diagonals.