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MathematicsCh-7 15 min comprehensive revision
NCERT Class 10 Mathematics — Chapter 7

Coordinate Geometry

Cartesian coordinates, Distance Formula derivation, Section Formula derivation (internal division), Mid-point formula, Centroid of a triangle, and collinearity.

Quick Key Takeaways:
Distance Formula: The Euclidean distance between P(x1,y1)P(x_1, y_1) and Q(x2,y2)Q(x_2, y_2) is: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
Distance from Origin: Distance of point P(x,y)P(x, y) from origin (0,0)(0, 0) is x2+y2\sqrt{x^2 + y^2}.
Section Formula: Coordinates of point P(x,y)P(x, y) dividing line segment A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2) internally in ratio m1:m2m_1 : m_2 are: x=m1x2+m2x1m1+m2,y=m1y2+m2y1m1+m2x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \quad y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}
Mid-point Formula: Special case where m1:m2=1:1m_1 : m_2 = 1 : 1: M=(x1+x22,y1+y22)M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)
Centroid of Triangle: For vertices (x1,y1),(x2,y2),(x3,y3)(x_1, y_1), (x_2, y_2), (x_3, y_3): G=(x1+x2+x33,y1+y2+y33)G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)
Graph & Coordinate VisualizerPlot coordinates, visualize distance formula, and explore section formula divisions
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Jump to Concept:1. 1. Cartesian Foundations & The Distance Formula Derivation2. 2. The Section Formula Derivation & Collinearity Principles3. 3. Important Solved Board Examination Questions (3-Mark & 5-Mark)4. 4. CBSE Case-Study Modeling: GPS Grid Navigation for Sports Day Track5. 5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric
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1. Cartesian Foundations & The Distance Formula Derivation

Axioms & Foundational Theory

Comprehensive mathematical foundations, definitions, formulas, and visual conceptual model for Coordinate Geometry.

• Derivation of the Distance Formula (Pythagorean Theorem)
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Let P(x1,y1)P(x_1, y_1) and Q(x2,y2)Q(x_2, y_2) be two points in the Cartesian plane.
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Draw perpendiculars PRPR and QSQS to the X-axis, and draw PT⊥QSPT \perp QS.
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In the right-angled triangle ΔPTQ\Delta PTQ:
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Base PT=RS=OS−OR=x2−x1PT = RS = OS - OR = x_2 - x_1
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Perpendicular QT=QS−TS=QS−PR=y2−y1QT = QS - TS = QS - PR = y_2 - y_1
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Applying Pythagoras Theorem in ΔPTQ\Delta PTQ:
PQ2=PT2+QT2=(x2−x1)2+(y2−y1)2PQ^2 = PT^2 + QT^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2
PQ=(x2−x1)2+(y2−y1)2\mathbf{PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}
📊 Coordinate Geometry: Distance & Section FormulasVisual Model
A(x₁, y₁)B(x₂, y₂)P(x, y)m₁m₂DISTANCE FORMULAd = √[(x₂ - x₁)² + (y₂ - y₁)²]SECTION FORMULA (P divides AB)x = (m₁x₂ + m₂x₁) / (m₁ + m₂)

Visual schematic mapping the Cartesian plane, Euclidean distance line segment, internal division section ratio, and the midpoint coordinates.

2

2. The Section Formula Derivation & Collinearity Principles

Theorem Proofs & Derivations

Step-by-step rigorous geometric and algebraic theorem proofs and formula derivations for Coordinate Geometry.

• Derivation of the Section Formula (Internal Division)
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Let P(x,y)P(x, y) divide A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2) internally in the ratio m1:m2m_1 : m_2, so that APPB=m1m2\frac{AP}{PB} = \frac{m_1}{m_2}.
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Drop perpendiculars to the X-axis from A,P,BA, P, B and draw horizontal lines to form similar right triangles ΔARP∼ΔPTB\Delta AR P \sim \Delta PTB:
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APPB=ARPT=PRBT\frac{AP}{PB} = \frac{AR}{PT} = \frac{PR}{BT}
m1m2=x−x1x2−x  ⟹  m1(x2−x)=m2(x−x1)  ⟹  (m1+m2)x=m1x2+m2x1\frac{m_1}{m_2} = \frac{x - x_1}{x_2 - x} \implies m_1(x_2 - x) = m_2(x - x_1) \implies (m_1 + m_2)x = m_1 x_2 + m_2 x_1
x=m1x2+m2x1m1+m2\mathbf{x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}}
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Similarly for the Y-coordinate: y=m1y2+m2y1m1+m2\mathbf{y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}}
• Trisection of a Line Segment
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Points of trisection divide the line segment in the ratios 1:21 : 2 and 2:12 : 1.
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3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

Topper Step Solutions

Standard CBSE board exam numericals and riders with complete step-by-step mathematical working and justifications.

• 3-Mark Standard Board Question: Find the ratio in which the Y-axis divides the line segment joining the points (5,−6)(5, -6) and (−1,−4)(-1, -4). Also find the point of intersection.
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Step 1 (Ratio Setup): Let the Y-axis divide the segment joining A(5,−6)A(5, -6) and B(−1,−4)B(-1, -4) in the ratio k:1k : 1 at point P(0,y)P(0, y) (since any point on Y-axis has X-coordinate =0= 0).
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Step 2 (Section Formula for X-coordinate):
x=k(x2)+1(x1)k+1x = \frac{k(x_2) + 1(x_1)}{k + 1}
0=k(−1)+1(5)k+1  ⟹  −k+5=0  ⟹  k=50 = \frac{k(-1) + 1(5)}{k + 1} \implies -k + 5 = 0 \implies \mathbf{k = 5}
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Step 3 (Ratio): The ratio is k:1=5:1k : 1 = \mathbf{5 : 1}.
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Step 4 (Y-coordinate Calculation):
y=k(y2)+1(y1)k+1=5(−4)+1(−6)5+1=−20−66=−266=−133y = \frac{k(y_2) + 1(y_1)}{k + 1} = \frac{5(-4) + 1(-6)}{5 + 1} = \frac{-20 - 6}{6} = \frac{-26}{6} = -\frac{13}{3}
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Final Boxed Answer: Ratio=5:1,Point of Intersection=(0,−133)\mathbf{\text{Ratio} = 5 : 1, \quad \text{Point of Intersection} = \left(0, -\frac{13}{3}\right)}
• 5-Mark Heavyweight Board Problem / Rider: Find the coordinates of the points of trisection of the line segment joining A(2,−2)A(2, -2) and B(−7,4)B(-7, 4).
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Step 1 (Trisection Points Setup): Let PP and QQ be the points of trisection of ABAB. Then PP divides ABAB in the ratio 1:21 : 2, and QQ divides ABAB in the ratio 2:12 : 1 (or QQ is the midpoint of PBPB).
•
Step 2 (Coordinates of PP with ratio 1:21 : 2):
xP=1(−7)+2(2)1+2=−7+43=−33=−1x_P = \frac{1(-7) + 2(2)}{1 + 2} = \frac{-7 + 4}{3} = \frac{-3}{3} = -1
yP=1(4)+2(−2)1+2=4−43=03=0y_P = \frac{1(4) + 2(-2)}{1 + 2} = \frac{4 - 4}{3} = \frac{0}{3} = 0
P=(−1,0)\mathbf{P = (-1, 0)}
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Step 3 (Coordinates of QQ using midpoint of PBPB):
xQ=xP+xB2=−1+(−7)2=−82=−4x_Q = \frac{x_P + x_B}{2} = \frac{-1 + (-7)}{2} = \frac{-8}{2} = -4
yQ=yP+yB2=0+42=42=2y_Q = \frac{y_P + y_B}{2} = \frac{0 + 4}{2} = \frac{4}{2} = 2
Q=(−4,2)\mathbf{Q = (-4, 2)}
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Final Boxed Answer: Points of Trisection are P(−1,0) and Q(−4,2)\mathbf{\text{Points of Trisection are } P(-1, 0) \text{ and } Q(-4, 2)}
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4. CBSE Case-Study Modeling: GPS Grid Navigation for Sports Day Track

Case Study Mastery (4 Marks)

Real-world mathematical modeling scenario with structured multi-part questions and full step solutions.

• Practical Application Context: GPS Grid Navigation for Sports Day Track
To conduct Sports Day activities in a rectangular school ground ABCDABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along ADAD. Niharika runs 14\frac{1}{4}th the distance ADAD on the 2nd line and posts a green flag. Preet runs 15\frac{1}{5}th the distance ADAD on the 8th line and posts a red flag.
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Q1: Find the coordinates of Niharika's green flag (NN) and Preet's red flag (PP). →\rightarrow Total AD=100AD = 100 m. Niharika's Y-coord =14×100=25= \frac{1}{4} \times 100 = 25 m on 2nd line   ⟹  N(2,25)\implies \mathbf{N(2, 25)}. Preet's Y-coord =15×100=20= \frac{1}{5} \times 100 = 20 m on 8th line   ⟹  P(8,20)\implies \mathbf{P(8, 20)}.
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Q2: Find the distance between both flags. →\rightarrow Distance NP=(8−2)2+(20−25)2=62+(−5)2=36+25=61 meters≈7.81NP = \sqrt{(8 - 2)^2 + (20 - 25)^2} = \sqrt{6^2 + (-5)^2} = \sqrt{36 + 25} = \mathbf{\sqrt{61}\text{ meters}} \approx 7.81 m.
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Q3: If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post it? →\rightarrow Midpoint formula: M=(2+82,25+202)=(102,452)=(5,22.5)M = \left(\frac{2 + 8}{2}, \frac{25 + 20}{2}\right) = \left(\frac{10}{2}, \frac{45}{2}\right) = \mathbf{(5, 22.5)}. She must post the blue flag on the 5th line at a distance of 22.5 meters.
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5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

Important Solved Board Questions

Examiner step-marking allocations, common algebraic/arithmetic deduction traps, and presentation guidelines.

• Step-by-Step Marking Rubric & Presentation Guidelines
•
1 Mark: Writing correct coordinate points from case context.
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1 Mark: Applying distance formula (x2−x1)2+(y2−y1)2\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}.
•
1 Mark: Section formula substitution m1:m2m_1:m_2.
•
1 Mark: Correct midpoint calculation with final physical units.
• Common Calculation Traps & Verification Checklist
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Trap 1: When asked for the ratio in which an axis divides a segment, assuming (x,0)(x, 0) for Y-axis (X-axis has y=0y = 0; Y-axis has x=0x = 0).
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Trap 2: Forgetting that distance is always non-negative (d≥0d \ge 0).
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Trap 3: Mixing up coordinates x1,x2x_1, x_2 and y1,y2y_1, y_2 in the Section formula.
Authentic Board Question (3 Marks)Topic: Coordinate Geometry Mathematical Proof & Computations
State the governing mathematical theorem/formula, show complete geometric or computational steps, and find the exact result for Coordinate Geometry.

Official CBSE Step-by-Step Marking Breakdown:

Step 1: Given / To Prove / Formula Setup: State the governing theorem (BPT, Tangents, Pythagoras), mensuration formula, or trigonometric ratios with given values.
1 Mark
Step 2: Step-by-Step Derivation / Arithmetic Working: Show complete deductive steps or numerical calculations with π=22/7\pi = 22/7 or trigonometric standard values.
1 Mark
Step 3: Final Boxed Answer with Units / Q.E.D.: State the final numerical result with proper square/cubic units (cm2,m3\text{cm}^2, \text{m}^3) or formal proof conclusion.
1 Mark
Model Student Answer (Target: Full 3/3 Marks):
To score full 3 marks on Coordinate Geometry in CBSE Mathematics:

1. Theorem / Formula: Write the standard governing formula or state the geometric theorem clearly.
2. Calculation / Proof: Substitute dimensions systematically or provide deductive step justifications.
3. Final Result: Box the final answer with required units (cm,cm2,m3\text{cm}, \text{cm}^2, \text{m}^3) or conclude with "Hence Proved".
Examiner Mark Deduction Traps:
•Always write intermediate calculation lines to earn partial credit under CBSE step-marking.
•State reasons in parentheses (e.g., "[Tangents from an external point are equal]") during geometric proofs.

High-Frequency Conceptual Doubts & FAQs

Curated answers to the most common questions asked by Class 10 students.
The distance between P(x1,y1)P(x_1, y_1) and Q(x2,y2)Q(x_2, y_2) is d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. It is derived by constructing a right triangle with horizontal base ∣x2−x1∣|x_2 - x_1| and vertical height ∣y2−y1∣|y_2 - y_1| and applying Pythagoras theorem: d2=(Δx)2+(Δy)2d^2 = (\Delta x)^2 + (\Delta y)^2.

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