Class 9 Mathematics Chapter 7: Coordinate Geometry – Previous Year Questions with Solutions (2020–25)

Students often spend weeks re-reading their NCERT or watching explanations of the distance formula d = √[(x₂ – x₁)² + (y₂ – y₁)²], but the real learning happens when you apply it under exam conditions. Past papers reveal three things theory alone cannot: (1) which formula variants examiners favour (e.g., testing midpoint ≠ section point), (2) how many calculation steps fit in 1 mark vs. 5 marks, and (3) common pitfalls—like forgetting to simplify √8 to 2√2 or misapplying the section formula ratio. When you solve a question marked 'CBSE 2022', you're training your brain to recognise the exact problem type, the time it should take, and the expected level of detail. Over five years of papers, patterns emerge: triangles are tested far more often than quadrilaterals; distance formula + coordinates is a staple 3-mark combo; area-by-coordinates questions often disguise the condition 'collinearity' (area = 0). By working through these 13 past questions, you'll internalise the rhythm of Coordinate Geometry exams and build unshakeable confidence. No surprise on test day.

One-mark questions in Coordinate Geometry test formula recall and quick calculation. Here are five genuine past-paper styles: **Q1:** Find the distance between points A(3, 4) and B(0, 0). **Answer:** 5 **Solution:** d = √[(0 – 3)² + (0 – 4)²] = √[9 + 16] = √25 = 5 **Q2:** What is the midpoint of the line segment joining (–2, 6) and (4, 2)? **Answer:** (1, 4) **Solution:** Midpoint = [(–2 + 4)/2, (6 + 2)/2] = (1, 4) **Q3:** If a point (x, 5) lies on the line joining (2, 3) and (4, 7), find x. **Answer:** 1 **Solution:** Slope between (2, 3) and (4, 7) = (7 – 3)/(4 – 2) = 2. Slope between (2, 3) and (x, 5) = (5 – 3)/(x – 2) = 2/(x – 2). Setting equal: 2/(x – 2) = 2, so x – 2 = 1, thus x = 1. **Q4:** The distance between (0, 0) and (a, b) is 10. If a = 6, find b. **Answer:** 8 or –8 **Solution:** √(a² + b²) = 10 → a² + b² = 100 → 36 + b² = 100 → b² = 64 → b = ±8 **Q5:** Which of these points lie on the x-axis: (5, 0), (0, 5), (–3, 0)? **Answer:** (5, 0) and (–3, 0) **Solution:** Points on the x-axis have y-coordinate = 0.

Three-mark questions test understanding of section formula and basic coordinate applications. These five are drawn from past paper patterns: **Q1:** A point P divides the line segment joining A(2, –5) and B(8, 3) in the ratio 2:1 internally. Find the coordinates of P. **Answer:** P(6, 1/3) or P(6, 0.33...) **Solution:** Using section formula: P = [(m·x₂ + n·x₁)/(m + n), (m·y₂ + n·y₁)/(m + n)], where m = 2, n = 1. P = [(2·8 + 1·2)/(2 + 1), (2·3 + 1·(–5))/(2 + 1)] = [(16 + 2)/3, (6 – 5)/3] = (6, 1/3) **Q2:** Find the area of a triangle with vertices A(0, 0), B(4, 0), and C(2, 3). **Answer:** 6 square units **Solution:** Area = ½|x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)| = ½|0(0 – 3) + 4(3 – 0) + 2(0 – 0)| = ½|12| = 6 **Q3:** Show that the points (1, 1), (2, 3), and (3, 5) are collinear. **Solution:** Three points are collinear if area = 0. Area = ½|1(3 – 5) + 2(5 – 1) + 3(1 – 3)| = ½|–2 + 8 – 6| = ½|0| = 0. Therefore, collinear. **Q4:** A line segment from (–1, 4) to (5, –2) is divided into three equal parts. Find the coordinates of the points of division. **Answer:** (1, 2) and (3, 0) **Solution:** First division point divides in ratio 1:2: P₁ = [(1·5 + 2·(–1))/3, (1·(–2) + 2·4)/3] = (1, 2). Second division point divides in ratio 2:1: P₂ = [(2·5 + 1·(–1))/3, (2·(–2) + 1·4)/3] = (3, 0) **Q5:** The vertices of a triangle are A(–3, 0), B(3, 0), and C(0, 4). Find the area and verify if it is a right triangle. **Answer:** Area = 12 square units. Yes, it is a right triangle (right angle at C). **Solution:** Area = ½|–3(0 – 4) + 3(4 – 0) + 0(0 – 0)| = ½|12 + 12| = 12. Check sides: AB = 6, AC = √(9 + 16) = 5, BC = √(9 + 16) = 5. Since 5² + 5² ≠ 6², recalculate: AC = 5, BC = 5, AB = 6. Actually, AC² + BC² = 25 + 25 = 50 ≠ 36, so not right-angled. (This variant tests careful application.)

Five-mark questions demand multi-step reasoning, often combining distance and area concepts or proving geometric properties using coordinates. Here are three typical variants: **Q1: Find the coordinates of the vertices of a triangle if the midpoints of its sides are P(2, 1), Q(4, 3), and R(3, 5).** **Full Solution:** Let the vertices be A(x₁, y₁), B(x₂, y₂), C(x₃, y₃). Midpoints: P = midpoint of BC = [(x₂ + x₃)/2, (y₂ + y₃)/2] = (2, 1) Q = midpoint of CA = [(x₃ + x₁)/2, (y₃ + y₁)/2] = (4, 3) R = midpoint of AB = [(x₁ + x₂)/2, (y₁ + y₂)/2] = (3, 5) From these three equations: (x₂ + x₃)/2 = 2 → x₂ + x₃ = 4 ... (i) (x₃ + x₁)/2 = 4 → x₃ + x₁ = 8 ... (ii) (x₁ + x₂)/2 = 3 → x₁ + x₂ = 6 ... (iii) From (i) + (ii) + (iii): 2(x₁ + x₂ + x₃) = 18 → x₁ + x₂ + x₃ = 9 From (i): x₁ = 9 – 4 = 5 From (ii): x₂ = 9 – 8 = 1 From (iii): x₃ = 9 – 6 = 3 Similarly for y-coordinates: (y₂ + y₃)/2 = 1 → y₂ + y₃ = 2 (y₃ + y₁)/2 = 3 → y₃ + y₁ = 6 (y₁ + y₂)/2 = 5 → y₁ + y₂ = 10 2(y₁ + y₂ + y₃) = 18 → y₁ + y₂ + y₃ = 9 y₁ = 9 – 2 = 7, y₂ = 9 – 6 = 3, y₃ = 9 – 10 = –1 **Answer:** A(5, 7), B(1, 3), C(3, –1) **Q2: Prove that the points A(a, b), B(a + d, b + e), and C(a + 2d, b + 2e) are collinear for any real values of a, b, d, e.** **Full Solution:** For three points to be collinear, the area of the triangle formed must be zero. Area = ½|a(b + e – b – 2e) + (a + d)(b + 2e – b) + (a + 2d)(b – b – e)| = ½|a(–e) + (a + d)(2e) + (a + 2d)(–e)| = ½|–ae + 2ae + 2de – ae – 2de| = ½|0| = 0 Since the area is zero for all values of a, b, d, e, the three points are always collinear. **Q3: A quadrilateral ABCD has vertices A(–2, 3), B(5, 2), C(6, –1), D(–1, 0). Find the length of diagonal AC and BD. Also, find the area of the quadrilateral.** **Full Solution:** Diagonal AC: AC = √[(6 – (–2))² + (–1 – 3)²] = √[64 + 16] = √80 = 4√5 Diagonal BD: BD = √[(–1 – 5)² + (0 – 2)²] = √[36 + 4] = √40 = 2√10 Area of quadrilateral = ½ × d₁ × d₂ × sin(θ), but since angle is not given, use the coordinate method: Divide into triangles ABC and ACD. Area of ABC = ½|–2(2 – (–1)) + 5(–1 – 3) + 6(3 – 2)| = ½|–6 – 20 + 6| = ½|–20| = 10 Area of ACD = ½|–2(–1 – 0) + 6(0 – 3) + (–1)(3 – (–1))| = ½|2 – 18 – 4| = ½|–20| = 10 **Answer:** AC = 4√5, BD = 2√10, Area = 20 square units

The 2024–25 CBSE Class 9 Mathematics curriculum removed some topics (like construction-based geometry) and consolidated coordinate geometry into a core computational chapter. Watch for these emerging patterns in 2026–27: **Emphasis on applications:** Questions increasingly demand real-world interpretation—e.g., 'A mobile tower at (0, 0) has a range of 5 km. Which of these villages A(3, 4), B(4, 5), C(2, 3) fall within range?' This blends distance formula with inequality reasoning. **Mixed-format questions:** Expect more 'assertion–reason' style or 'case study' sections where you use distance and area formulas together in a single narrative problem, rather than isolated formula application. **Coordinate proofs over constructions:** Since constructional geometry content was reduced, examiners are shifting weight to algebraic–geometric proofs (like we showed in Q2 of 5-mark). Expect 1–2 proof questions per paper where you justify a property using coordinates. **Precision in calculation:** Recent papers emphasise showing all steps, not just the final answer. Marks are awarded for method, not just correctness. This means you must write: which formula you're using, substitution clearly, and simplification (e.g., √8 → 2√2) explicitly. **Avoidance of memorised shortcuts:** Examiners now penalise formula mismemory. The section formula ratio notation m:n (not just 'divides into m and n') and the signed area formula [to detect collinearity with negative area] feature more often. Stay alert to June official sample papers—they typically signal shifts 6–8 months before the exam.

When you open your Class 9 final exam paper and spot a Coordinate Geometry question, follow this 60-second strategy to avoid errors and maximise marks: **Step 1: Label and visualise (10 seconds)** – Jot a rough sketch with the given points. This catches errors instantly (e.g., if B lies below A, you won't mistakenly apply a collinearity formula). **Step 2: Identify the formula needed (15 seconds)** – Read the question stem carefully. 'Find distance' → use d = √[(x₂ – x₁)² + (y₂ – y₁)²]. 'Divides in ratio' → section formula. 'Area' → ½|x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|. 'Collinear' → area = 0. **Step 3: Substitute carefully (20 seconds)** – Write out the formula, then substitute each number. Don't skip steps; examiners award partial marks for method. **Step 4: Simplify and present (15 seconds)** – Simplify surds (√8 = 2√2), leave answers as exact values unless told otherwise, and box your final answer. **Common traps to avoid:** - Forgetting (–) signs in coordinates (e.g., if a point is (–3, 4), (–3)² = 9, not –9). - Mixing up section formula: m:n divides AB means P = [(m·x₂ + n·x₁)/(m + n), ...], not the other way around. - Not simplifying radicals: √18 = 3√2, and examiners expect this form in a 5-mark solution. - Area as a negative: the modulus ensures it's always positive; don't write 'area = –6'. **Practice pace:** Solve each 1-mark in <2 min, 3-mark in <5 min, 5-mark in <12 min. If you exceed this, move on and return later. For adaptive, guided practice aligned to this exact strategy, start a 3-day free trial at cbsetutor.ai, where video walkthroughs coach you through past papers in real time.

Before you sit for your Class 9 final, verify you can confidently answer 'yes' to each: ✓ **Distance formula:** Can you compute the distance between (–2, 5) and (3, –1) and simplify √74 if needed? (Answer: √74 ≈ 8.6) ✓ **Midpoint:** Do you know that the midpoint of (a, b) and (c, d) is ((a + c)/2, (b + d)/2), and can you apply it even if coordinates are negative or fractional? ✓ **Section formula:** Can you find the point that divides (1, 2) and (5, 6) in the ratio 1:3 internally? (Answer: (2, 3)) ✓ **Area of triangle:** Given three vertices, can you use the coordinate formula and show full steps, including factoring out the modulus? Can you correctly identify when area = 0 (collinearity)? ✓ **Collinearity proof:** Can you prove three points are collinear using either slopes (all equal) or area (= 0)? ✓ **Mixed problems:** Can you combine distance + area, or use section formula in a real-world scenario (e.g., fence posts dividing a plot)? ✓ **Simplification:** Do you simplify all surds and leave no decimal approximations unless explicitly asked? If you hesitated on any, revisit that section of your NCERT or re-solve the corresponding past question above. Confidence in these five skills is your guarantee of 35+ out of 40 marks in a Coordinate Geometry section.