Chapter 3 of 24

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ICSE Class 10 Mathematics★ ICSE Class 10 Mathematics · ~25 marks · Board: ICSE

AP, GP, Coordinate Geometry, Reflection & Equation of a Line

ARITHMETIC and GEOMETRIC PROGRESSIONS — sequences that grow by ADDING and MULTIPLYING. COORDINATE GEOMETRY — reflection, section formula, midpoint, and the EQUATION OF A LINE. ICSE Class 10 Mathematics.

⏱ 18 min readMedium difficulty✓ Free · NCERT-aligned

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Study Guide

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Key points, glossary and the questions most likely to show up.

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By the end of this chapter you'll be able to…

1Find the nth term and sum of n terms of both arithmetic and geometric progressions; find the sum to infinity of a GP when |r| < 1
2Apply reflection rules to reflect points and shapes in the x-axis, y-axis, origin, and the line y = x
3Apply the section formula (internal division) to find coordinates of a point dividing a segment in a given ratio; find midpoints and centroids
4Write equations of straight lines in slope-intercept, point-slope, two-point, and intercept forms; determine if lines are parallel or perpendicular
5Find the distance of a point from a given line; write the equation of a line through a given point parallel or perpendicular to a given line

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Why this chapter matters

AP and GP are guaranteed questions in the ICSE board — at least 4-6 marks per paper, and the formulas are simple to memorise and apply. Coordinate geometry (reflection, section formula, equation of a line) accounts for another 8-10 marks. The equation of a line is the most-applied coordinate concept across Class 10 and ISC Class 11-12. ICSE consistently tests: 'Find the equation of the line through a point, parallel/perpendicular to a given line.' Students who master the slope-intercept and point-slope forms can answer these questions in under 5 minutes.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Progressions, Coordinate Geometry & Lines

1. Arithmetic Progression (AP)

nth Term: aₙ = a + (n—1)d

a = first term. d = common difference.

Sum of First n Terms: Sₙ = (n/2)[2a + (n—1)d] = (n/2)(a + l)

l = last term.

2. Geometric Progression (GP)

nth Term: aₙ = a × rⁿ⁻¹

r = common ratio = a₂/a₁ = a₃/a₂ = ...

Sum of First n Terms

For r > 1: Sₙ = a(rⁿ — 1) / (r — 1)
For r < 1: Sₙ = a(1 — rⁿ) / (1 — r)

Sum of INFINITE GP (|r| < 1): S∞ = a / (1 — r)

3. Reflection (Mirroring Points)

Reflection in	Point (x, y) becomes
x-axis	(x, —y)
y-axis	(—x, y)
Origin	(—x, —y)
Line y = x	(y, x)

4. Section and Midpoint Formula

Distance: d = √[(x₂—x₁)² + (y₂—y₁)²]

Section Formula (Internal Division in ratio m:n)

((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n))

Midpoint: ((x₁+x₂)/2, (y₁+y₂)/2)

Centroid of Triangle

G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)

5. Equation of a Line

Slope (m)

m = (y₂—y₁)/(x₂—x₁) = tan θ. Angle from positive x-axis.

Forms

Form	Equation	When to Use
Slope-Intercept	y = mx + c	Given slope and y-intercept
Point-Slope	y — y₁ = m(x — x₁)	Given one point and slope
Two-Point	y — y₁ = [(y₂—y₁)/(x₂—x₁)](x — x₁)	Given two points
Intercept	x/a + y/b = 1	Given intercepts

Parallel and Perpendicular Lines

Parallel: m₁ = m₂
Perpendicular: m₁ × m₂ = —1

Distance of a Point from a Line

d = |Ax₁ + By₁ + C| / √(A² + B²) (for line Ax + By + C = 0)

Equation from Two Conditions

ICSE frequently asks: 'Find the equation of a line passing through P and parallel/perpendicular to a given line.' → Find slope from the given line. Apply parallel/perpendicular condition. Use point-slope form. Simplify.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Arithmetic Progression (AP)

GENERAL TERM: aₙ = a + (n−1)d (a = first term, d = common difference). SUM of n terms: Sₙ = (n/2)[2a + (n−1)d] = (n/2)(a + l) where l = last term. TO FIND d: d = a₂ − a₁ = a₃ − a₂ (constant difference). CHECKING AP: Three numbers a, b, c are in AP iff b − a = c − b, i.e., 2b = a + c (b is arithmetic mean of a and c).

COMMON ICSE TRICK: If the question asks for 3 terms in AP, take them as (a−d), a, (a+d). Sum = 3a. Product involves d. For 4 terms: (a−3d), (a−d), (a+d), (a+3d). This makes algebraic manipulation much easier than using a, a+d, a+2d, a+3d.

Geometric Progression (GP)

GENERAL TERM: aₙ = a × rⁿ⁻¹ (a = first term, r = common ratio). SUM of n terms: Sₙ = a(rⁿ−1)/(r−1) for r > 1. Sₙ = a(1−rⁿ)/(1−r) for r < 1. SUM TO INFINITY: S∞ = a/(1−r) ONLY when |r| < 1. TO FIND r: r = a₂/a₁ = a₃/a₂ (constant ratio). THREE terms in GP: take them as a/r, a, ar — product = a³, much simpler.

S∞ only exists when |r| < 1 — the series CONVERGES. If |r| ≥ 1, there is no finite sum to infinity. For ICSE: if the question gives a fraction ratio (e.g., 1/2), always check |r| < 1 before applying S∞.

Reflection Rules

REFLECTION IN x-axis: (x, y) → (x, −y). REFLECTION IN y-axis: (x, y) → (−x, y). REFLECTION IN ORIGIN: (x, y) → (−x, −y). REFLECTION IN LINE y = x: (x, y) → (y, x). REFLECTION IN LINE y = −x: (x, y) → (−y, −x). INVARIANT POINTS: Points that do not change under reflection. For x-axis: any point on x-axis (y=0). For y-axis: any point on y-axis (x=0). For y=x: any point where y=x (i.e., on the line itself).

ICSE frequently asks: 'The point P(3, 4) is reflected in the x-axis to give P′. P′ is then reflected in the y-axis to give P″. Find P″.' Answer: P → P′ = (3, −4) [x-axis]. P′ → P″ = (−3, −4) [y-axis]. Note: reflecting in both axes sequentially = reflecting in the origin.

Section Formula and Equation of a Line

DISTANCE FORMULA: d = √[(x₂−x₁)² + (y₂−y₁)²]. SECTION FORMULA (internal division in m:n): P = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)). MIDPOINT: M = ((x₁+x₂)/2, (y₁+y₂)/2). CENTROID of triangle: G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3). SLOPE: m = (y₂−y₁)/(x₂−x₁) = tan θ. LINE FORMS: y = mx + c (slope-intercept). y − y₁ = m(x − x₁) (point-slope). x/a + y/b = 1 (intercept form). PARALLEL: m₁ = m₂. PERPENDICULAR: m₁ × m₂ = −1. DISTANCE OF POINT (x₁,y₁) from line Ax+By+C=0: |Ax₁+By₁+C|/√(A²+B²).

MOST COMMON ICSE LINE QUESTION: 'Find the equation of a line passing through A(2,3) and perpendicular to the line 2x − 3y + 5 = 0.' STEPS: (1) Find slope of given line: 2x−3y+5=0 → y = (2/3)x + 5/3, so m₁ = 2/3. (2) Perpendicular slope: m₂ = −3/2 (negative reciprocal). (3) Point-slope form: y−3 = −3/2(x−2). Simplify.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Applying S∞ = a/(1-r) when |r| ≥ 1

✓ S∞ = a/(1−r) ONLY applies when |r| < 1. If r = 2 (or any value where |r| ≥ 1), the GP DIVERGES — terms get larger and larger, and there is no finite sum. ICSE questions on GP sum to infinity will always give a ratio where |r| < 1 (typically a fraction like 1/2, 1/3, −1/4). Always verify: is |r| < 1? If not, do not apply S∞.

WATCH OUT

✗ Confusing section formula numerator arrangement (which coordinate gets m, which gets n)

✓ Section formula for point dividing A(x₁,y₁) and B(x₂,y₂) in ratio m:n: x-coordinate = (m×x₂ + n×x₁)/(m+n). The SECOND point (B, ratio m) gets multiplied by m; the FIRST point (A, ratio n) gets multiplied by n. Memory aid: 'ratio m:n → m × B coordinate + n × A coordinate.' A common error is reversing this. Always: the ratio closer to the second point goes with the second point's coordinates.

WATCH OUT

✗ Forgetting to use the negative reciprocal for perpendicular slope

✓ If line 1 has slope m₁, a PERPENDICULAR line has slope m₂ = −1/m₁ (negative reciprocal). Two steps: (1) take the reciprocal (flip the fraction), (2) change the sign. Example: if m₁ = 2/3, then m₂ = −3/2. If m₁ = −4, then m₂ = 1/4. PARALLEL lines have the SAME slope (m₁ = m₂). PERPENDICULAR: m₁ × m₂ = −1 (the product is always −1). A horizontal line (m = 0) is perpendicular to a vertical line (undefined slope) — this is a special case.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· AP

The 5th term of an AP is 18 and the 9th term is 30. Find (a) the common difference, (b) the first term, (c) the 15th term.

▶ Show solution

Using aₙ = a + (n−1)d: a₅ = a + 4d = 18 ...(1). a₉ = a + 8d = 30 ...(2). Subtracting (1) from (2): 4d = 12, so d = 3. From (1): a + 4(3) = 18 → a + 12 = 18 → a = 6. (c) a₁₅ = a + 14d = 6 + 14(3) = 6 + 42 = 48. Answers: (a) d = 3, (b) a = 6, (c) a₁₅ = 48. Verify: a₅ = 6+4(3) = 18 ✓. a₉ = 6+8(3) = 30 ✓.

Q2MEDIUM· coordinate-section-formula

Find the coordinates of the point that divides the line segment joining A(−3, 5) and B(9, −7) internally in the ratio 1:2.

▶ Show solution

Using section formula (internal division in ratio m:n = 1:2): x = (m×x₂ + n×x₁)/(m+n) = (1×9 + 2×(−3))/(1+2) = (9 − 6)/3 = 3/3 = 1. y = (m×y₂ + n×y₁)/(m+n) = (1×(−7) + 2×5)/(1+2) = (−7 + 10)/3 = 3/3 = 1. The point is P(1, 1). Verification: AP = √[(1−(−3))² + (1−5)²] = √[16+16] = 4√2. PB = √[(9−1)² + (−7−1)²] = √[64+64] = 8√2. AP:PB = 4√2 : 8√2 = 1:2 ✓.

Q3HARD· GP-sum-infinity

The sum of the first 3 terms of a GP is 21 and the product of these terms is 216. Find the GP.

▶ Show solution

Let the three terms be a/r, a, ar. Sum: a/r + a + ar = 21 ...(1). Product: (a/r) × a × ar = a³ = 216 → a = 6. Substituting a = 6 into (1): 6/r + 6 + 6r = 21 → 6/r + 6r = 15 → Multiply by r: 6 + 6r² = 15r → 6r² − 15r + 6 = 0 → 2r² − 5r + 2 = 0 → (2r − 1)(r − 2) = 0. r = 1/2 or r = 2. If r = 2: terms are 6/2, 6, 6×2 = 3, 6, 12. GP: 3, 6, 12. If r = 1/2: terms are 6/(1/2), 6, 6×(1/2) = 12, 6, 3. GP: 12, 6, 3 (same terms in reverse). Both are valid GPs. Check: 3+6+12 = 21 ✓. 3×6×12 = 216 ✓.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•AP: aₙ = a+(n-1)d. Sₙ = n/2[2a+(n-1)d]. 3 terms in AP: take a-d, a, a+d.
•GP: aₙ = arⁿ⁻¹. Sₙ = a(rⁿ-1)/(r-1). S∞ = a/(1-r) only when |r|<1.
•3 terms in GP: a/r, a, ar. Product = a³ (very useful for problems).
•Reflection: x-axis → (x,-y). y-axis → (-x,y). Origin → (-x,-y). y=x → (y,x).
•Section formula (m:n): x = (mx₂+nx₁)/(m+n). y = (my₂+ny₁)/(m+n).
•Midpoint: ((x₁+x₂)/2, (y₁+y₂)/2). Centroid: ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3).
•Slope: m = (y₂-y₁)/(x₂-x₁). Parallel: m₁=m₂. Perpendicular: m₁×m₂=-1.
•Point-slope form: y-y₁ = m(x-x₁). Slope-intercept: y=mx+c.
•Distance of point from line |Ax₁+By₁+C|/√(A²+B²) — always use absolute value.
•Common mistake: in perpendicular, take NEGATIVE RECIPROCAL of original slope.

ICSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Research the AM-GM-HM inequality — for positive numbers, Arithmetic Mean ≥ Geometric Mean ≥ Harmonic Mean, with equality only when all numbers are equal. This single inequality unlocks many olympiad problems. Investigate its proof using induction and its applications in optimisation problems.
Investigate Zeno's paradox and its resolution through infinite geometric series — the sum 1/2 + 1/4 + 1/8 + ... = 1 shows that infinitely many terms can have a finite sum (when |r| < 1, S∞ = a/(1−r)). This insight led directly to the development of calculus in the 17th century.
Explore the Apollonius circle locus problem — given two fixed points A and B, the locus of points P satisfying PA/PB = k (constant ≠ 1) is a circle. The section formula and distance formula together prove this. This generalises the perpendicular-bisector idea (k = 1).
Research the role of reflections in symmetry groups — every regular polygon has a finite symmetry group consisting of rotations and reflections (the dihedral group). The square's symmetry group D4 has 8 elements (4 rotations + 4 reflections). This is the gateway from school geometry to abstract algebra.

Where else this chapter is tested

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

METHOD: Use the TWO-POINT FORM: y − y₁ = [(y₂ − y₁)/(x₂ − x₁)] × (x − x₁). Step-by-step: (1) Calculate the slope m = (y₂−y₁)/(x₂−x₁). (2) Use point-slope form with EITHER of the two points: y − y₁ = m(x − x₁). (3) Expand and simplify to standard form. Example: Find the equation of the line through A(1, 3) and B(4, 9). Slope = (9−3)/(4−1) = 6/3 = 2. Using A(1,3): y − 3 = 2(x − 1) → y − 3 = 2x − 2 → y = 2x + 1. VERIFY: Does B(4,9) satisfy y = 2x + 1? 2(4) + 1 = 9 ✓.

A point is INVARIANT under a reflection if it MAPS TO ITSELF — its image is the same as the original point. For reflection in the X-AXIS: a point (x, y) maps to (x, −y). For it to be invariant: (x, y) = (x, −y) → y = −y → y = 0. So ALL POINTS ON THE X-AXIS are invariant under reflection in the x-axis. For reflection in the Y-AXIS: invariant points have x = 0 (the y-axis itself). For reflection in y = x: invariant points have (x, y) = (y, x) → x = y. All points ON THE LINE y = x are invariant. For reflection in the origin: invariant only at (0, 0). ICSE question type: 'Which points are invariant when reflecting in the line y = x?' Answer: all points where y = x (the diagonal line).

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