Exercise 7.2 — Section Formula

Section formula and centroid of the triangle.

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Exercise 7.2 — Section Formula, Midpoint, Trisection & Centroid

Exercise 7.2 is the heart of Chapter 7, Coordinate Geometry, in Class 10 Mathematics (CBSE, Telangana & AP Board). It introduces four closely related tools — the Section Formula, the Midpoint Formula, Trisection Points, and the Centroid of a Triangle — all of which describe how a line segment or triangle can be divided or measured in the coordinate plane.

This exercise contains 10 problems covering: finding the point that divides a segment in a given ratio, finding trisection points, working backwards to find the ratio from a given dividing point, parallelogram and circle problems using midpoint, dividing into four equal parts, algebraic coordinates, and centroid calculations for three different triangles.

Section Formula Midpoint Formula Trisection Points Division into Equal Parts Centroid of Triangle

💡 Why this exercise matters: The Section Formula and Centroid appear in every Telangana SSC & AP SSC board exam — usually as a 4-mark or 5-mark question. Mastering the pattern of substituting m, n, x₁, y₁, x₂, y₂ correctly is all you need to score full marks.

Key Formulas — Section Formula, Midpoint, Trisection & Centroid

📐 Section Formula (Internal Division)

Point P dividing A(x₁,y₁)–B(x₂,y₂) in ratio m:n

P = ( (mx₂ + nx₁)/(m+n) , (my₂ + ny₁)/(m+n) )

📍 Midpoint Formula

Special case of section formula when m = n (ratio 1:1)

M = ( (x₁+x₂)/2 , (y₁+y₂)/2 )

✂️ Trisection Points

P divides AB in 1:2, Q divides AB in 2:1

P = ( (2x₁+x₂)/3 , (2y₁+y₂)/3 ) Q = ( (x₁+2x₂)/3 , (y₁+2y₂)/3 )

△ Centroid of Triangle

For triangle with vertices A, B, C

G = ( (x₁+x₂+x₃)/3 , (y₁+y₂+y₃)/3 )

Visual Guide — Section Formula & Trisection

Section Formula — P divides AB in ratio m:n

Trisection — P and Q divide AB into three equal parts

📌 Derivation hint (for board exams): The section formula is proved using the AA similarity of triangles △PAQ and △BPC, formed by drawing perpendiculars from A, P, and B to the x-axis. The ratio AP:PB = m:n gives the proportion AQ/PC = PQ/BC = m/n, from which x and y are solved. Know this 3-4 line derivation — it appears as a 2-mark "prove" question.

Questions 1–5: Section Formula & Midpoint Applications

Find the point dividing (−1, 7) and (4, −3) in ratio 2 : 3

A = (−1, 7), B = (4, −3), m = 2, n = 3. Apply section formula directly.

x = (mx₂ + nx₁)/(m+n) = (2×4 + 3×(−1))/(2+3) = (8 − 3)/5 = 5/5 = 1 y = (my₂ + ny₁)/(m+n) = (2×(−3) + 3×7)/(2+3) = (−6 + 21)/5 = 15/5 = 3 ∴ P = (1, 3)

(1, 3)

Find the trisection points of (4, −1) and (−2, −3)

A = (4, −1), B = (−2, −3). P divides AB in 1:2 (nearer to A), Q divides AB in 2:1 (nearer to B).

Point P = ((2x₁+x₂)/3, (2y₁+y₂)/3) x = (2×4 + (−2))/3 = (8−2)/3 = 6/3 = 2 y = (2×(−1) + (−3))/3 = (−2−3)/3 = −5/3 P = (2, −5/3) Point Q = ((x₁+2x₂)/3, (y₁+2y₂)/3) x = (4 + 2×(−2))/3 = (4−4)/3 = 0/3 = 0 y = (−1 + 2×(−3))/3 = (−1−6)/3 = −7/3 Q = (0, −7/3)

P = (2, −5/3) and Q = (0, −7/3)

Find the ratio in which (−1, 6) divides the segment joining (−3, 10) and (6, −8)

Let P(−1, 6) divide AB in ratio m:n. Set up section formula and compare x-coordinates.

Setup (6m + (−3)n)/(m+n) = −1 6m − 3n = −1(m + n) = −m − n 6m + m = −n + 3n 7m = 2n → m/n = 2/7 → m:n = 2:7 Verify y-coordinate: (−8×2 + 10×7)/(2+7) = (−16+70)/9 = 54/9 = 6 ✓ ∴ Point (−1, 6) divides the segment in ratio 2 : 7

✅ Tip: Always verify the answer using the y-coordinate equation after solving from x. If both match, the solution is confirmed. Examiners award a mark specifically for this verification step.

Ratio = 2 : 7

ABCD is a parallelogram with A(1,2), B(4,y), C(x,6), D(3,5). Find x and y.

Key property: Diagonals of a parallelogram bisect each other. So midpoint of AC = midpoint of BD.

Midpt AC = ((1+x)/2, (2+6)/2) = ((1+x)/2, 4) Midpt BD = ((4+3)/2, (y+5)/2) = (7/2, (y+5)/2) x-coords: (1+x)/2 = 7/2 → 1+x = 7 → x = 6 y-coords: 4 = (y+5)/2 → y+5 = 8 → y = 3 ∴ x = 6, y = 3

x = 6, y = 3

AB is diameter of circle. Centre = (2, −3), B = (1, 4). Find A.

The centre of a circle is the midpoint of any diameter. So midpoint of AB = Centre (2, −3).

Setup Let A = (x, y). Midpoint of AB = ((x+1)/2, (y+4)/2) = (2, −3) x: (x+1)/2 = 2 → x+1 = 4 → x = 3 y: (y+4)/2 = −3 → y+4 = −6 → y = −10 ∴ A = (3, −10)

A = (3, −10)

Questions 6–9: Division into Parts, Four Equal Parts & Algebraic Coordinates

A = (−2, −2), B = (2, −4). Find P on AB such that AP = (3/7)AB.

First convert the length condition into a ratio. AP/AB = 3/7 → AP = 3 parts, AB = 7 parts → PB = 7−3 = 4 parts.

Ratio AP : PB = 3 : 4 (since AP = 3/7 of total, PB = 4/7 of total) Apply Section formula with m=3, n=4 x = (3×2 + 4×(−2))/(3+4) = (6−8)/7 = −2/7 y = (3×(−4) + 4×(−2))/(3+4) = (−12−8)/7 = −20/7 ∴ P = (−2/7, −20/7)

📌 Converting length ratios to section ratios: When given AP = (3/7)AB, divide both sides by PB = AB − AP = (4/7)AB. Then AP:PB = (3/7):(4/7) = 3:4. This conversion step is commonly tested in board exams.

P = (−2/7, −20/7)

Divide A(−4, 0) and B(0, 6) into four equal parts. Find P, Q, R.

Strategy: Find Q = midpoint of AB first, then P = midpoint of AQ, and R = midpoint of QB. Three midpoint calculations.

Step 1 Q = midpoint of AB = ((−4+0)/2, (0+6)/2) = (−2, 3) Step 2 P = midpoint of A(−4,0) and Q(−2,3) = ((−4+(−2))/2, (0+3)/2) = (−6/2, 3/2) = (−3, 3/2) Step 3 R = midpoint of Q(−2,3) and B(0,6) = ((−2+0)/2, (3+6)/2) = (−1, 9/2) ∴ P = (−3, 3/2), Q = (−2, 3), R = (−1, 9/2)

P=(−3, 3/2), Q=(−2, 3), R=(−1, 9/2)

Divide A(−2, 2) and B(2, 8) into four equal parts. Find P, Q, R.

Same three-midpoint strategy as Q7.

Step 1 Q = midpoint of AB = ((−2+2)/2, (2+8)/2) = (0, 5) Step 2 P = midpoint of A(−2,2) and Q(0,5) = ((−2+0)/2, (2+5)/2) = (−1, 7/2) Step 3 R = midpoint of Q(0,5) and B(2,8) = ((0+2)/2, (5+8)/2) = (1, 13/2) ∴ P = (−1, 7/2), Q = (0, 5), R = (1, 13/2)

P=(−1, 7/2), Q=(0, 5), R=(1, 13/2)

Divide A(a+b, a−b) and B(a−b, a+b) in ratio 3 : 2 internally

Algebraic coordinates — apply section formula with m = 3, n = 2 and expand carefully.

x = (3(a−b) + 2(a+b)) / (3+2) = (3a − 3b + 2a + 2b) / 5 = (5a − b) / 5 y = (3(a+b) + 2(a−b)) / (3+2) = (3a + 3b + 2a − 2b) / 5 = (5a + b) / 5 ∴ P = ((5a − b)/5, (5a + b)/5)

((5a−b)/5, (5a+b)/5)

Question 10 — Centroid of Triangles

The centroid G of a triangle is the point where all three medians meet. A median connects a vertex to the midpoint of the opposite side. The centroid always divides each median in the ratio 2:1 (two-thirds from vertex to midpoint). Its coordinates are simply the average of the three vertices' coordinates.

Centroid Formula: G = ( (x₁+x₂+x₃)/3 , (y₁+y₂+y₃)/3 )

Q10 (i)

Centroid of triangle with vertices (−1, 3), (6, −3) and (−3, 6)

G = ((−1 + 6 + (−3))/3, (3 + (−3) + 6)/3) = ((−1 + 6 − 3)/3, (3 − 3 + 6)/3) = (2/3, 6/3) ∴ G = (2/3, 2)

(2/3, 2)

Q10 (ii)

Centroid of triangle with vertices (6, 2), (0, 0) and (4, −7)

G = ((6 + 0 + 4)/3, (2 + 0 + (−7))/3) = (10/3, (2 − 7)/3) = (10/3, −5/3) ∴ G = (10/3, −5/3)

(10/3, −5/3)

Q10 (iii)

Centroid of triangle with vertices (1, −1), (0, 6) and (−3, 0)

G = ((1 + 0 + (−3))/3, (−1 + 6 + 0)/3) = ((1 + 0 − 3)/3, (−1 + 6)/3) = (−2/3, 5/3) ∴ G = (−2/3, 5/3)

(−2/3, 5/3)

Quick Reference — All Answers at a Glance

Q#	Problem	Formula Used	Answer
Q1	(−1,7) & (4,−3), ratio 2:3	Section Formula	(1, 3)
Q2	Trisection of (4,−1) & (−2,−3)	Trisection	(2, −5/3) and (0, −7/3)
Q3	Ratio for point (−1,6) on (−3,10)–(6,−8)	Reverse Section	2 : 7
Q4	Parallelogram ABCD, find x and y	Midpoint (diagonals)	x=6, y=3
Q5	Diameter AB, centre (2,−3), B=(1,4)	Midpoint Formula	A = (3, −10)
Q6	AP = (3/7)AB on A(−2,−2)–B(2,−4)	Section (ratio 3:4)	(−2/7, −20/7)
Q7	4 equal parts of A(−4,0)–B(0,6)	Midpoint ×3	(−3, 3/2), (−2, 3), (−1, 9/2)
Q8	4 equal parts of A(−2,2)–B(2,8)	Midpoint ×3	(−1, 7/2), (0, 5), (1, 13/2)
Q9	(a+b,a−b) & (a−b,a+b), ratio 3:2	Section Formula	((5a−b)/5, (5a+b)/5)
Q10(i)	Centroid of (−1,3),(6,−3),(−3,6)	Centroid	(2/3, 2)
Q10(ii)	Centroid of (6,2),(0,0),(4,−7)	Centroid	(10/3, −5/3)
Q10(iii)	Centroid of (1,−1),(0,6),(−3,0)	Centroid	(−2/3, 5/3)

Common Mistakes to Avoid

Swapping m and n with x₁ and x₂: The section formula is (mx₂ + nx₁)/(m+n) — m always goes with x₂ (the second point's coordinate) and n with x₁ (the first). Swapping them gives the wrong point. Always match m with B's coordinates and n with A's.
Wrong ratio direction in trisection: P (closer to A) divides in ratio 1:2, and Q (closer to B) divides in ratio 2:1. Students often reverse this — always draw a quick sketch to confirm which point is closer to which end.
Forgetting to verify with y in Q3-type problems: When finding the ratio from a given dividing point, always verify using the y-coordinate equation after finding m:n from the x-equation. In Q3, both give 2:7 — but one might differ if there's a calculation error.
Wrong diagonal pairing in parallelograms: In Q4, the diagonals are AC and BD (not AB and CD). Diagonals connect opposite vertices. In ABCD, A connects to C, and B connects to D.
Not simplifying AP:PB from AP/AB: In Q6, AP = (3/7)AB means AP/AB = 3/7 but AP:PB = 3:4 (not 3:7). The ratio is between the two parts, not each part to the whole.
Sign errors with negative coordinates: In Q10(i), −1 + 6 + (−3) = 2, not −8. Write out every addition step explicitly; board examiners check intermediate working.

⚠️ Board Exam Alert (Telangana & AP SSC): The most frequently tested problem type from this exercise is the reverse section formula (like Q3 — "find the ratio in which a given point divides…"). The key trap is that students set up the formula for x but forget to verify using y, losing a mark. Always write: "Verification: y-coordinate → [calculation] = [given y] ✓" as the last line of these problems.

What This Exercise Prepares You For

Exercise 7.2 is the central toolkit of Chapter 7. The section formula and midpoint formula are applied again in Exercise 7.3 (area of triangles using coordinates) and are the basis for all coordinate proofs in geometry. The centroid formula directly connects to the properties of medians studied in Chapter 7 of Class 9.

For students who need to review the distance formula (used to verify results like Q3 in some approaches), see Exercise 7.1. The algebraic coordinate techniques from Q9 also prepare students for parametric problems in Class 11 coordinate geometry.

📐 Board Exam Tip (Telangana & AP SSC): For every section formula problem, write four things clearly: (1) identify A(x₁,y₁) and B(x₂,y₂), (2) state m and n, (3) write the formula, (4) substitute and simplify. Examiners follow this four-step format when awarding partial marks — even a wrong final answer can score 2 out of 4 marks if steps 1–3 are written correctly.