Curriculum
Mathematics Grade 10
Grade 10 Mathematics
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Chapter 1. Real Numbers
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Chapter 2. Polynomials
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Chapter 3: PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
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Chapter 4: QUADRATIC EQUATIONS
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Chapter 5: ARITHMETIC PROGRESSIONS
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Chapter 6: TRIANGLES
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Chapter 7: COORDINATE GEOMETRY
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Chapter 8: INTRODUCTION TO TRIGONOMETRY
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Chapter 9: SOME APPLICATIONS OF TRIGONOMETRY
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Chapter 10: CIRCLES
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Chapter 11: AREAS RELATED TO CIRCLES
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Chapter 12: SURFACE AREAS AND VOLUMES
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Chapter 13: STATISTICS
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Chapter 14: PROBABILITY
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Interactive Worksheets
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Chapter Wise Worksheets
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Solved Problems and solutions
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Chapter 7: COORDINATE GEOMETRY
Grade 10 · Mathematics · Chapter 7
Coordinate Geometry
Find the exact distance between any two points, locate where a segment splits in any ratio, and pinpoint midpoints — all using just two coordinates.
| Distance Formula | Section Formula | Mid-Point Formula | Applications |
What You Will Learn
| ✦ Recap of coordinate axes, abscissa and ordinate | ✦ Deriving the Distance Formula using Pythagoras |
| ✦ Distance from the origin: OP = √(x²+y²) | ✦ Testing collinearity, shape of triangles & quadrilaterals |
| ✦ Section Formula — internal division | ✦ Mid-point Formula as a special case (ratio 1:1) |
| ✦ Worked Examples 1–16 (fully solved) | ✦ Practice Sets A–D with collapsible answers |
| ✦ Real-life applications — relay towers, flags, maps | ✦ Chapter Summary and 8-point exam quick-check |
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Section 1 Introduction — The Coordinate Plane |
To locate any point on a plane we use two perpendicular number lines — the x-axis (horizontal) and the y-axis (vertical) — meeting at the origin O(0,0). Every point P in the plane has a unique address (x, y) called its coordinates.
Quick Recap — Key Vocabulary
| Abscissa (x-coordinate) | Signed distance of the point from the y-axis. Positive to the right, negative to the left. |
| Ordinate (y-coordinate) | Signed distance of the point from the x-axis. Positive above, negative below. |
| Point on x-axis | Always of the form (x, 0) — ordinate is zero. |
| Point on y-axis | Always of the form (0, y) — abscissa is zero. |
| Origin | (0, 0) — both coordinates are zero. |
Visual — Coordinate Plane (4 Quadrants)
| Quadrant II (−x, +y) |
Quadrant I (+x, +y) |
| Quadrant III (−x, −y) |
Quadrant IV (+x, −y) |
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Section 2 The Distance Formula |
Before deriving the general formula, let us build intuition with simpler cases.
Building Up — From Simple to General
| Both points on x-axis | A(4,0) and B(6,0): distance = |6 − 4| = 2 units. Simply subtract x-coordinates. |
| Both points on y-axis | C(0,3) and D(0,8): distance = |8 − 3| = 5 units. Simply subtract y-coordinates. |
| One on each axis | A(4,0) and C(0,3): Use Pythagoras. AC = √(4²+3²) = √25 = 5 units. |
| General: any two points | Derive using Pythagoras — see below. |
Derivation of the Distance Formula
Given P(x₁, y₁) and Q(x₂, y₂). Draw PR ⊥ x-axis and QS ⊥ x-axis. Draw PT ⊥ QS.
Then: PT = x₂ − x₁ and QT = y₂ − y₁
In right-angled △PTQ: PQ² = PT² + QT² = (x₂−x₁)² + (y₂−y₁)²
∴ PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]
Distance Formula
PQ = √ [ (x₂ − x₁)² + (y₂ − y₁)² ]
Distance from origin O(0,0): OP = √(x² + y²)
Diagram — Distance Formula Derivation
| P(x₁,y₁) | PT = x₂−x₁ | Q(x₂,y₂) |
| T | QT = y₂−y₁ |
WORKED EXAMPLE 1 — Type of triangle from three points
Do points P(3,2), Q(−2,−3) and R(2,3) form a triangle? If so, what type?
QR = √[(−2−2)² + (−3−3)²] = √[16+36] = √52 ≈ 7.21
PR = √[(3−2)² + (2−3)²] = √[1+1] = √2 ≈ 1.41
Sum of any two sides > third side ⇒ triangle exists.
Check: PQ² + PR² = 50 + 2 = 52 = QR² ⇒ by converse of Pythagoras, ∠P = 90°
PQR is a right-angled triangle. ✓
WORKED EXAMPLE 2 — Prove a quadrilateral is a square
Show that A(1,7), B(4,2), C(−1,−1), D(−4,4) are vertices of a square.
BC = √[(4+1)²+(2+1)²] = √[25+9] = √34
CD = √[(−1+4)²+(−1−4)²] = √[9+25] = √34
DA = √[(1+4)²+(7−4)²] = √[25+9] = √34
Diagonal AC = √[(1+1)²+(7+1)²] = √[4+64] = √68
Diagonal BD = √[(4+4)²+(2−4)²] = √[64+4] = √68
All four sides equal AND both diagonals equal. ABCD is a square. ✓
WORKED EXAMPLE 3 — Are three students seated in a line?
A(3,1), B(6,4) and C(8,6) are three seats in a classroom grid. Are they collinear?
BC = √[(8−6)²+(6−4)²] = √[4+4] = √8 = 2√2
AC = √[(8−3)²+(6−1)²] = √[25+25] = √50 = 5√2
AB + BC = 3√2 + 2√2 = 5√2 = AC ⇒ A, B, C are collinear. Yes, they sit in a line. ✓
WORKED EXAMPLE 4 — Find a relation: equidistant point
Find the relation between x and y if point P(x,y) is equidistant from A(7,1) and B(3,5).
(x−7)² + (y−1)² = (x−3)² + (y−5)²
x²−14x+49 + y²−2y+1 = x²−6x+9 + y²−10y+25
−14x + 49 − 2y + 1 = −6x + 9 − 10y + 25
−8x + 8y = −16 ⇒ x − y = 2
This is the perpendicular bisector of AB — any equidistant point lies on it.
WORKED EXAMPLE 5 — Find a point on y-axis equidistant from two points
Find the point on the y-axis equidistant from A(6,5) and B(−4,3).
PA² = PB² ⇒ (6−0)² + (5−y)² = (−4−0)² + (3−y)²
36 + 25 − 10y + y² = 16 + 9 − 6y + y²
61 − 10y = 25 − 6y ⇒ 36 = 4y ⇒ y = 9
Required point = (0, 9). ✓
Verify: PA = √(36+16) = √52, PB = √(16+36) = √52 ✓
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Section 3 The Section Formula |
Suppose you need to find the exact coordinates of a point P that lies on segment AB and divides it in a given ratio. The Section Formula gives the answer directly.
Real-Life Motivation — Relay Tower Problem
Town A is at origin O. Town B is at (36, 15). A relay tower P must be placed so that PB = 2 × PA, i.e., P divides AB in the ratio 1:2. Using proportional reasoning with similar triangles, we find P = (12, 5). The Section Formula generalises this idea to any ratio.
Derivation of the Section Formula
Let P(x,y) divide A(x₁,y₁) and B(x₂,y₂) internally in ratio m₁:m₂.
Draw AR, PS, BT ⊥ x-axis and AQ, PC ∥ x-axis. By AA similarity: △PAQ ~ △BPC
∴ PA/BP = AQ/PC = PQ/BC
AQ = x − x₁ | PC = x₂ − x | PQ = y − y₁ | BC = y₂ − y
m₁/m₂ = (x−x₁)/(x₂−x) ⇒ m₁(x₂−x) = m₂(x−x₁)
Solving: x = (m₁x₂ + m₂x₁) / (m₁ + m₂)
Similarly: y = (m₁y₂ + m₂y₁) / (m₁ + m₂)
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Section Formula (Internal Division) P = ( (m₁x₂+m₂x₁)/(m₁+m₂) , (m₁y₂+m₂y₁)/(m₁+m₂) ) |
Mid-Point Formula (ratio 1:1) M = ( (x₁+x₂)/2 , (y₁+y₂)/2 ) |
When ratio is k:1 (useful shortcut)
P = ( (kx₂+x₁)/(k+1) , (ky₂+y₁)/(k+1) )
Assume ratio k:1, solve for k, then find coordinates.
WORKED EXAMPLE 6 — Direct use of Section Formula
Find the coordinates of the point dividing (4,−3) and (8,5) in the ratio 3:1 internally.
x = (3×8 + 1×4)/(3+1) = (24+4)/4 = 28/4 = 7
y = (3×5 + 1×(−3))/(3+1) = (15−3)/4 = 12/4 = 3
Required point = (7, 3). ✓
WORKED EXAMPLE 7 — Find the ratio given the dividing point
In what ratio does (−4, 6) divide the segment joining A(−6,10) and B(3,−8)?
−4 = (k×3 + 1×(−6))/(k+1) = (3k−6)/(k+1)
−4(k+1) = 3k−6 ⇒ −4k−4 = 3k−6 ⇒ 2 = 7k ⇒ k = 2/7
Ratio = k:1 = 2/7 : 1 = 2:7
Verify with y-coordinate: (2×(−8) + 7×10)/(2+7) = (−16+70)/9 = 54/9 = 6 ✓
(−4, 6) divides AB in ratio 2:7. ✓
WORKED EXAMPLE 8 — Points of trisection
Find the coordinates of the points of trisection of segment joining A(2,−2) and B(−7,4).
P (ratio 1:2):
x = (1×(−7) + 2×2)/(1+2) = (−7+4)/3 = −3/3 = −1
y = (1×4 + 2×(−2))/(1+2) = (4−4)/3 = 0
P = (−1, 0)
Q (ratio 2:1):
x = (2×(−7) + 1×2)/(2+1) = (−14+2)/3 = −12/3 = −4
y = (2×4 + 1×(−2))/(2+1) = (8−2)/3 = 2
Q = (−4, 2)
Points of trisection: (−1, 0) and (−4, 2). ✓
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P = (−1, 0)
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Q = (−4, 2)
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WORKED EXAMPLE 9 — y-axis divides a segment
Find the ratio in which the y-axis divides (5,−6) and (−1,−4). Find the point of intersection.
x = (k×(−1) + 1×5)/(k+1) = 0 ⇒ −k+5 = 0 ⇒ k = 5
Ratio = 5:1
y = (5×(−4) + 1×(−6))/(5+1) = (−20−6)/6 = −26/6 = −13/3
y-axis divides the segment in ratio 5:1 and point of intersection = (0, −13/3). ✓
WORKED EXAMPLE 10 — Missing vertex of parallelogram
A(6,1), B(8,2), C(9,4), D(p,3) are vertices of a parallelogram. Find p.
Mid-point of AC = ((6+9)/2, (1+4)/2) = (15/2, 5/2)
Mid-point of BD = ((8+p)/2, (2+3)/2) = ((8+p)/2, 5/2)
Equate x-coordinates: 15/2 = (8+p)/2 ⇒ 15 = 8+p ⇒ p = 7
D = (7, 3). ✓
WORKED EXAMPLE 11 — Sports Day Flag Problem
Niharika plants a flag at 1/4 of AD on the 2nd column ⇒ N(2,25). Preet plants at 1/5 of AD on 8th column ⇒ P(8,20). Find distance NP and midpoint (Rashmi’s flag position).
N = (2, 25), P = (8, 20)
NP = √[(8−2)² + (20−25)²] = √[36+25] = √61 m ≈ 7.81 m
Rashmi’s flag (midpoint): x = (2+8)/2 = 5, y = (25+20)/2 = 22.5
Rashmi plants at the 5th column, 22.5 m from A. ✓
WORKED EXAMPLE 12 — Centroid of a triangle
Find the centroid of the triangle with vertices A(4,−1), B(−2,−3), C(0,7).
G = ( (x₁+x₂+x₃)/3 , (y₁+y₂+y₃)/3 )
G = ( (4+(−2)+0)/3 , (−1+(−3)+7)/3 ) = ( 2/3 , 3/3 ) = (2/3, 1) ✓
WORKED EXAMPLE 13 — Divide in ratio, then find
A and B are (−2,−2) and (2,−4). Find P such that AP = (3/7)AB and P lies on AB.
P = ( (3×2 + 4×(−2))/(3+4) , (3×(−4) + 4×(−2))/(3+4) )
x = (6−8)/7 = −2/7 y = (−12−8)/7 = −20/7
P = (−2/7, −20/7). ✓
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Section 4 Formula Reference Card |
| Formula | Expression | When to Use |
|---|---|---|
| Distance Formula | PQ = √[(x₂−x₁)²+(y₂−y₁)²] | Find length of any segment |
| Distance from Origin | OP = √(x²+y²) | When one point is (0,0) |
| Section Formula | P = ((m₁x₂+m₂x₁)/(m₁+m₂), (m₁y₂+m₂y₁)/(m₁+m₂)) | Point divides AB in ratio m₁:m₂ |
| Mid-Point Formula | M = ((x₁+x₂)/2, (y₁+y₂)/2) | Mid-point of any segment |
| Collinearity Test | AB + BC = AC (or area of triangle = 0) | Check 3 points lie on a line |
| Centroid | G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3) | Centre of gravity of triangle |
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Section 5 Practice Exercises with Answers |
Set A — Distance Formula
Use PQ = √[(x₂−x₁)²+(y₂−y₁)²]. Show all working.
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(1) Find the distance between: (i) (2,3) and (4,1) (ii) (−5,7) and (−1,3) (iii) (a,b) and (−a,−b)
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(2) Find the distance between (0,0) and (36,15). What does this represent geometrically?
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(3) Are (1,5), (2,3) and (−2,−11) collinear?
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(4) Check whether (5,−2), (6,4) and (7,−2) are vertices of an isosceles triangle.
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(5) Find the point on x-axis equidistant from (2,−5) and (−2,9).
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(6) Find the values of y for which PQ = 10, where P(2,−3) and Q(10,y).
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Reveal Answers ▼
(2) √(1296+225)=√1521=39 units (distance between towns A and B)
(3) AB=√5, BC=√160, AC=√245. AB+BC ≠ AC — not collinear
(4) AB=√37, BC=√40, CA=√37 ⇒ AB=CA ⇒ isosceles triangle
(5) Let point = (x,0). Equate distances ⇒ (−7, 0)
(6) (10−2)²+(y+3)²=100 ⇒ (y+3)²=36 ⇒ y+3=±6 ⇒ y=3 or y=−9
Set B — Types of Triangles and Quadrilaterals
Compute all sides and diagonals where needed. State the type with reasons.
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(1) Name the quadrilateral formed by (−1,−2), (1,0), (−1,2), (−3,0). Give reasons.
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(2) Name the quadrilateral formed by (4,5), (7,6), (4,3), (1,2).
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(3) ABCD has A(3,0), B(4,5), C(−1,4), D(−2,−1). Find all sides and diagonals. What type?
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(4) If Q(0,1) is equidistant from P(5,−3) and R(x,6), find x and distances QR and PR.
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(5) Find the relation between x and y so that (x,y) is equidistant from (3,6) and (−3,4).
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(6) 4 friends sit at A(3,4), B(6,7), C(−1,4), D(6,1). Champa says ABCD is a square. Chameli disagrees. Who is correct?
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Reveal Answers ▼
(2) AB=BC=CD=DA=√10 and diagonals AC=√8≠BD=√40 ⇒ rhombus (not square)
(3) All sides = √26, diagonals = √32 and √32 ⇒ equal sides and equal diagonals ⇒ square
(4) QP²=25+16=41; QR²=(x²+25)=41 ⇒ x=±4; QR=√41; PR = √[(5−x)²+81]
(5) 36+y²−12y+x²−6x = x²+6x+9+y²−8y+16 ⇒ 3x − 2y = 6
(6) AB=√18, BC=√50, CD=√50, DA=√18 — sides not all equal ⇒ Chameli is correct, not a square
Set C — Section Formula and Mid-Point
Use the Section Formula and Mid-Point Formula. Show all steps.
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(1) Find coordinates of point dividing (−1,7) and (4,−3) in ratio 2:3.
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(2) Find trisection points of the segment joining (4,−1) and (−2,−3).
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(3) Find the ratio in which (−1,6) divides the segment joining (−3,10) and (6,−8).
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(4) Find the ratio in which the x-axis divides A(1,−5) and B(−4,5). Find point of division.
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(5) (1,2), (4,y), (x,6) and (3,5) are vertices of a parallelogram. Find x and y.
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(6) AB is a diameter of circle with centre (2,−3) and B=(1,4). Find A.
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Reveal Answers ▼
(2) P(ratio 1:2) = (2, −5/3); Q(ratio 2:1) = (0, −7/3)
(3) k:1 method ⇒ k=2/7 ⇒ ratio 2:7
(4) On x-axis y=0: 0=(k×5+1×(−5))/(k+1) ⇒ k=1; ratio=1:1; point = (−3/2, 0)
(5) Diagonals bisect ⇒ mid(AC)=mid(BD): (1+x)/2=(4+3)/2 ⇒ x=6; (2+6)/2=(y+5)/2 ⇒ y=3
(6) Centre is midpoint: (2,−3) = ((Ax+1)/2,(Ay+4)/2) ⇒ Ax=3, Ay=−10 ⇒ A = (3,−10)
Set D — Mixed and Challenge Problems
Combine distance, section and mid-point concepts. Verify answers.
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(1) A(−2,2) and B(2,8). Find the coordinates of the points dividing AB into four equal parts.
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(2) Find the area of a rhombus whose vertices are (3,0), (4,5), (−1,4), (−2,−1). [Hint: Area = ½ × d₁ × d₂]
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(3) P(1,2), Q(−3,−4) and R(5,−6). M is midpoint of PQ. Find the length of median RM.
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(4) ABCD is a rectangle with A(−1,2), B(3,−1), C(x,y), D. AB ∥ DC and AD ∥ BC. Find C and D.
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(5) The vertices of a triangle are (1,−1), (−4,6) and (−3,−5). Find the length of each median.
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(6) Town A is 36 km east of town B. Town C is 15 km north of town B. Find the distance from A to C directly.
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Reveal Answers ▼
(2) d₁ = dist (3,0)−(−1,4) = √32; d₂ = dist (4,5)−(−2,−1) = √72; Area = ½×√32×√72 = ½×√2304 = 24 sq units
(3) M=midpoint PQ=(−1,−1); RM=√[(5+1)²+(−6+1)²]=√[36+25]=√61
(4) AB vector=(4,−3); C = B+AD direction = B+AB = (3+4,−1−3) = (7,−4); D=A+(4,−3)=(3,−1) wait — D=(−1+4,2−3)=(3,−1)
(5) Find midpoints of each side, compute distance from opposite vertex to each midpoint
(6) AC = √(36²+15²) = √(1296+225) = √1521 = 39 km
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Section 6 Chapter Summary |
All Key Results at a Glance
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Distance Formula PQ = √[(x₂−x₁)²+(y₂−y₁)²]. From origin: OP = √(x²+y²). Order of points does not matter. |
Collinearity Three points A, B, C are collinear if and only if AB + BC = AC. Equivalently, area of triangle ABC = 0. |
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Section Formula P dividing AB in m₁:m₂: x=(m₁x₂+m₂x₁)/(m₁+m₂), y=(m₁y₂+m₂y₁)/(m₁+m₂) |
Mid-Point Formula M = ((x₁+x₂)/2, (y₁+y₂)/2). Special case of Section Formula when m₁=m₂=1. |
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Parallelogram Property Diagonals of a parallelogram bisect each other ⇒ mid-point of AC = mid-point of BD. Use this to find missing vertices. |
Centroid G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3). The centroid divides each median in ratio 2:1 from vertex. |
8-Point Quick-Check Before Your Exam
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1Distance is always non-negative — always take the positive square root.
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2For the Section Formula: m₁ goes with x₂/y₂ (far end), m₂ goes with x₁/y₁ (near end). Don’t mix these up.
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3When finding a ratio (k:1 trick), always verify the answer using the other coordinate too.
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4A point on the x-axis is (x, 0). A point on the y-axis is (0, y). These constraints simplify many problems.
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5To prove a quadrilateral is a square: show all 4 sides equal AND both diagonals equal. For a rhombus: all sides equal but diagonals unequal.
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6Trisection: first point divides in ratio 1:2, second point divides in ratio 2:1. Apply Section Formula twice.
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7For parallelogram vertex problems, use: mid-point of diagonal 1 = mid-point of diagonal 2. This always gives two equations in two unknowns.
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8The centroid formula G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3) is a direct extension of the mid-point idea — average of all three coordinates.
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Grade 10 Maths Chapter 7 — Coordinate Geometry
Topics covered: coordinate axes, abscissa, ordinate, quadrants, distance formula derivation using Pythagoras theorem, distance from origin, collinearity of three points, identifying triangles and quadrilaterals from vertices, equidistant points, Section Formula for internal division derivation, mid-point formula, trisection of a line segment, parallelogram diagonal property, centroid of a triangle, real-life applications — relay tower, sports day flag problem. CBSE Class 10 Maths Chapter 7 board exam revision.