✦ NCF 2023 • COMPETENCY-FOCUSED QUESTION BANK ✦
Chapter 7: Coordinate Geometry
Class 10 Standard Mathematics (Case Study Bank)
Curated By Akash Srivastva
✨ Interactive TLMs, Worksheets & Digital Study Notes:
www.akashmaths.online
Case Study 1
Precision Agriculture
4 Marks
Precision Agriculture (Sensors & Midpoint)
In smart agriculture, two IoT soil moisture sensors A(1, 3) and B(5, 7) are mapped in meters on a 2D Cartesian grid plane:
• Sensor A: (1, 3)
• Sensor B: (5, 7)
• Sampling Point M: Midpoint of AB
Q1. Distance Evaluation
1 Mark
Calculate the exact linear distance (in meters) between Sensor A and Sensor B.
(A) 6 meters
(B) 4√2 meters (≈ 5.66 m)
(C) 5√2 meters
(D) 8 meters
Q2. Midpoint Location
1 Mark
Find the coordinates of the midpoint M where the drone drops the soil-sampler.
(A) (3, 5)
(B) (2, 4)
(C) (4, 6)
(D) (3, 4)
Q3. Equidistance Property
2 Marks
If receiver station C(x, 1) is equidistant from Sensor A(1, 3) and midpoint M(3, 5), find x.
(A) x = 3
(B) x = 4
(C) x = 5
(D) x = 6
OR (Alternative Q3)
Find the coordinates of safety waypoint P lying on the y-axis equidistant from A(1, 3) and B(5, 7).
(A) (0, 8)
(B) (0, 6)
(C) (0, 7)
(D) (0, 9)
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | AB = √[(5 − 1)² + (7 − 3)²] = √[16 + 16] = √32 = 4√2 meters. | [0.5 Mark] for distance formula. [0.5 Mark] for 4√2 m. |
| Q2 | M = ((1 + 5)/2, (3 + 7)/2) = (6/2, 10/2) = (3, 5). | [1 Mark] for midpoint (3, 5). |
| Q3 | AC² = MC² ⇒ (x − 1)² + (1 − 3)² = (x − 3)² + (1 − 5)². x² − 2x + 5 = x² − 6x + 25 ⇒ 4x = 20 ⇒ x = 5. |
[1 Mark] for expanding equation. [1 Mark] for solving x = 5. |
| Q3 (OR) | Let P be (0, y). PA² = PB² ⇒ (0 − 1)² + (y − 3)² = (0 − 5)² + (y − 7)². 1 + y² − 6y + 9 = 25 + y² − 14y + 49 ⇒ 8y = 64 ⇒ y = 8 ⇒ P(0, 8). |
[1 Mark] for setting (0, y) and equating. [1 Mark] for P(0, 8). |
Case Study 2
Warehouse Automation
4 Marks
Warehouse AGV Route & Handover
AGV-1 at P(−2, −3) and AGV-2 at Q(6, 5) move along straight path PQ. A designated handover docking point R divides PQ in the ratio 3 : 1, and the path intersects the y-axis at S.
Q1. Separation Distance
1 Mark
Find the exact distance between AGV-1 at P(−2, −3) and AGV-2 at Q(6, 5).
(A) 10 units
(B) 8√2 units (≈ 11.31)
(C) 12 units
(D) 6√2 units
Q2. Handover Point
1 Mark
If point R divides line segment PQ internally in ratio 3 : 1, calculate coordinates of R.
(A) (4, 3)
(B) (3, 4)
(C) (2, 1)
(D) (5, 4)
Q3. Axis Intersection Ratio
2 Marks
Find the ratio in which the y-axis divides segment PQ and find intersection point S.
(A) 1 : 2; S(0, −2)
(B) 2 : 3; S(0, 1)
(C) 1 : 3; S(0, −1)
(D) 3 : 1; S(0, −1)
OR (Alternative Q3)
If sensor T is on segment PQ such that PT = (3/8)PQ, find coordinates of point T.
(A) (2, 0)
(B) (0, 1)
(C) (1, 1)
(D) (1, 0)
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | PQ = √[(6 − (−2))² + (5 − (−3))²] = √[8² + 8²] = √128 = 8√2 units. | [1 Mark] for distance 8√2 units. |
| Q2 | x = [3(6) + 1(−2)]/(3 + 1) = 16/4 = 4; y = [3(5) + 1(−3)]/(3 + 1) = 12/4 = 3 ⇒ R(4, 3). | [1 Mark] for coordinates R(4, 3). |
| Q3 | Let y-axis intersect at (0, y) in ratio k : 1 ⇒ 0 = [k(6) + 1(−2)]/(k + 1) ⇒ 6k = 2 ⇒ k = 1/3 (ratio 1 : 3). y = [(1/3)(5) − 3]/[(1/3) + 1] = (−4/3)/(4/3) = −1 ⇒ S(0, −1). |
[1 Mark] for ratio 1 : 3. [1 Mark] for point S(0, −1). |
| Q3 (OR) | PT : TQ = 3 : 5. x = [3(6) + 5(−2)]/8 = 8/8 = 1; y = [3(5) + 5(−3)]/8 = 0/8 = 0 ⇒ T(1, 0). |
[1 Mark] for ratio 3 : 5. [1 Mark] for point T(1, 0). |
Case Study 3
Sports Tracking
4 Marks
Football Pass & Midpoint
A football tactical analysis tool plots positions relative to pitch center spot O(0, 0):
• Midfielder M: (−4, −2)
• Striker S: (8, 6)
• Pass midpoint K: Center of passing vector MS
Q1. Distance to Origin
1 Mark
Calculate the straight-line distance of striker S(8, 6) from center spot O(0, 0).
(A) 10 units
(B) 14 units
(C) 12 units
(D) 8√2 units
Q2. Passing Midpoint
1 Mark
Find the coordinates of midpoint K along the ground pass between M(−4, −2) and S(8, 6).
(A) (4, 4)
(B) (2, 4)
(C) (2, 2)
(D) (1, 2)
Q3. Interception Points
2 Marks
Defender D stands on segment MS such that MD : DS = 1 : 3. Find coordinates of Defender D.
(A) (0, 1)
(B) (−1, 0)
(C) (−2, 0)
(D) (0, 0)
OR (Alternative Q3)
If pass deflects to W(x, y) such that point (1, 3) is midpoint of MW, find coordinates of W.
(A) (6, 8)
(B) (5, 7)
(C) (6, 6)
(D) (4, 8)
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | OS = √(8² + 6²) = √(64 + 36) = √100 = 10 units. | [1 Mark] for OS = 10 units. |
| Q2 | K = ((−4 + 8)/2, (−2 + 6)/2) = (4/2, 4/2) = (2, 2). | [1 Mark] for midpoint (2, 2). |
| Q3 | x = [1(8) + 3(−4)]/(1 + 3) = −4/4 = −1; y = [1(6) + 3(−2)]/4 = 0/4 = 0 ⇒ D(−1, 0). | [1 Mark] for section formula. [1 Mark] for D(−1, 0). |
| Q3 (OR) | (−4 + x)/2 = 1 ⇒ x = 6; (−2 + y)/2 = 3 ⇒ y = 8 ⇒ W(6, 8). | [1 Mark] for midpoint setup. [1 Mark] for W(6, 8). |
Case Study 4
Aerospace Radar
4 Marks
Satellite & Ground Stations
Ground stations A(−3, 5) and B(9, −7) track a high-altitude satellite P(x, y). Position parameters are calibrated when P is equidistant from both stations.
Q1. Baseline Distance
1 Mark
Find the exact distance between Ground Station A(−3, 5) and Ground Station B(9, −7).
(A) 16 units
(B) 12√2 units (≈ 16.97)
(C) 15 units
(D) 10√2 units
Q2. Equidistant Coordinate
1 Mark
If satellite P(5, y) is equidistant from Station A and Station B, find vertical coordinate y.
(A) y = 1
(B) y = 2
(C) y = 0
(D) y = −1
Q3. Calibration Nodes
2 Marks
Find coordinates of calibration node Q dividing segment AB internally in ratio 1 : 2.
(A) (2, 0)
(B) (0, 2)
(C) (1, 1)
(D) (3, −1)
OR (Alternative Q3)
Show whether midpoint of AB equals midpoint of alternative monitoring nodes C(10, 3) and D(−4, −5).
(A) Identical; both share midpoint (3, −1)
(B) Different; M_AB=(3, −1) but M_CD=(2, −1)
(C) Different; M_AB=(2, 1) but M_CD=(3, −1)
(D) Identical; both share midpoint (1, 1)
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | AB = √[(9 + 3)² + (−7 − 5)²] = √[144 + 144] = √288 = 12√2 units. | [1 Mark] for evaluating 12√2 units. |
| Q2 | PA² = PB² ⇒ (5 + 3)² + (y − 5)² = (5 − 9)² + (y + 7)². 64 + y² − 10y + 25 = 16 + y² + 14y + 49 ⇒ 24y = 24 ⇒ y = 1. |
[1 Mark] for y = 1. |
| Q3 | x = [1(9) + 2(−3)]/3 = 3/3 = 1; y = [1(−7) + 2(5)]/3 = 3/3 = 1 ⇒ Q(1, 1). | [1 Mark] for section formula. [1 Mark] for Q(1, 1). |
| Q3 (OR) | M_AB = ((−3 + 9)/2, (5 − 7)/2) = (3, −1). M_CD = ((10 − 4)/2, (3 − 5)/2) = (3, −1). Both midpoints are identical. |
[1 Mark] for computing both midpoints. [1 Mark] for proving equivalence (3, −1). |
Case Study 5
Smart Grid
4 Marks
Hybrid Substation (Ratio 3:1)
A smart electrical grid connects Wind Turbine T₁(−2, −5) and Solar Farm T₂(6, 7) in km units. Central substation P divides T₁T₂ internally in ratio 3 : 1.
Q1. Midpoint Verification
1 Mark
State the section formula and calculate the midpoint coordinates of line T₁T₂.
(A) (4, 2)
(B) (2, 1)
(C) (3, 2)
(D) (1, 2)
Q2. Substation Coordinates
1 Mark
Calculate the coordinates of Hybrid Substation P dividing T₁T₂ internally in ratio 3 : 1.
(A) (4, 4)
(B) (3, 3)
(C) (4, 3)
(D) (5, 5)
Q3. Grid Distances & Equidistance
2 Marks
Calculate the total distance (in km) between Wind Turbine T₁ and Solar Farm T₂.
(A) 12 km
(B) 16 km
(C) 4√13 km (≈ 14.42 km)
(D) 15 km
OR (Alternative Q3)
If generator C(2, k) is equidistant from T₁(−2, −5) and T₂(6, 7), find k.
(A) k = 2
(B) k = −1
(C) k = 0
(D) k = 1
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | Section Formula: ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)). Midpoint = ((−2 + 6)/2, (−5 + 7)/2) = (4/2, 2/2) = (2, 1). |
[0.5 Mark] for formula. [0.5 Mark] for midpoint (2, 1). |
| Q2 | x = [3(6) + 1(−2)]/4 = 16/4 = 4; y = [3(7) + 1(−5)]/4 = 16/4 = 4 ⇒ P(4, 4). | [1 Mark] for P(4, 4). |
| Q3 | d = √[(6 + 2)² + (7 + 5)²] = √[64 + 144] = √208 = 4√13 ≈ 14.42 km. | [1 Mark] for distance setup. [1 Mark] for 4√13 km. |
| Q3 (OR) | CT₁² = CT₂² ⇒ (2 + 2)² + (k + 5)² = (2 − 6)² + (k − 7)². 16 + k² + 10k + 25 = 16 + k² − 14k + 49 ⇒ 24k = 24 ⇒ k = 1. |
[1 Mark] for equating squared distances. [1 Mark] for solving k = 1. |
Case Study 6
Logistics Centroid
4 Marks
Centroid Sorting Hub
Three demand centers A(2, −3), B(−1, 5), and C(8, 4) form a triangle (1 grid unit = 10 km). Sorting hub G is positioned at the geometric centroid.
Q1. Centroid Formula
1 Mark
Calculate the centroid G(X, Y) coordinates of △ABC.
(A) (4, 3)
(B) (3, 2)
(C) (3, 3)
(D) (2, 2)
Q2. Ground Scale Distance
1 Mark
Calculate actual ground distance from hub G(3, 2) to Demand Center A(2, −3) where 1 unit = 10 km.
(A) 10√26 km (≈ 50.99 km)
(B) 50 km
(C) 5√26 km
(D) 60 km
Q3. Geometric Classification
2 Marks
Determine if △ABC is isosceles, scalene, or equilateral using side lengths.
(A) Equilateral (all sides equal)
(B) Isosceles (two sides equal)
(C) Scalene (AB = √73, BC = √82, CA = √85; all different)
(D) Right-angled isosceles
OR (Alternative Q3)
Find coordinates of satellite terminal T dividing AB in ratio 1 : 2 internally.
(A) (1, −1/3)
(B) (1, 1/3)
(C) (0, 1)
(D) (2, −1)
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | X = (2 − 1 + 8)/3 = 9/3 = 3; Y = (−3 + 5 + 4)/3 = 6/3 = 2 ⇒ G(3, 2). | [0.5 Mark] for formula. [0.5 Mark] for G(3, 2). |
| Q2 | GA = √[(2 − 3)² + (−3 − 2)²] = √[1 + 25] = √26 units. Ground distance = 10 × √26 ≈ 50.99 km. |
[0.5 Mark] for √26 units. [0.5 Mark] for 10√26 km. |
| Q3 | AB = √[(-3)² + 8²] = √73; BC = √[9² + (-1)²] = √82; CA = √[(-6)² + (-7)²] = √85. Since AB ≠ BC ≠ CA, it is a scalene triangle. |
[1 Mark] for calculating all three sides. [1 Mark] for scalene conclusion. |
| Q3 (OR) | x = [1(−1) + 2(2)]/3 = 3/3 = 1; y = [1(5) + 2(−3)]/3 = −1/3 ⇒ T(1, −1/3). | [1 Mark] for section formula. [1 Mark] for T(1, −1/3). |
Case Study 7
Archaeology Mapping
4 Marks
Trisection of Ancient Canal
Excavation points A(−3, 6) and B(6, −3) are joined by an ancient canal. Test wells P and Q trisect AB (AP = PQ = QB):
• Granary A: (−3, 6)
• Citadel B: (6, −3)
• Trisection: AP = PQ = QB
Q1. Trisection Ratios
1 Mark
State internal ratios in which P and Q divide segment AB.
(A) P divides in 1 : 2, and Q divides in 2 : 1
(B) Both divide in 1 : 1
(C) P divides in 1 : 3, and Q divides in 3 : 1
(D) P divides in 2 : 3, and Q divides in 3 : 2
Q2. First Trisection Well
1 Mark
Calculate coordinates of test well P closer to A(−3, 6).
(A) (1, 2)
(B) (0, 3)
(C) (3, 0)
(D) (−1, 4)
Q3. Well Coordinates & Chamber
2 Marks
Calculate the coordinates of the second test well Q closer to B(6, −3).
(A) (2, 1)
(B) (0, 3)
(C) (3, 0)
(D) (4, −1)
OR (Alternative Q3)
If P is the midpoint of AR where A(−3, 6), find coordinates of chamber R(x, y).
(A) (0, 3)
(B) (6, −3)
(C) (3, 3)
(D) (3, 0)
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | P divides AB in ratio 1 : 2, and Q divides AB in ratio 2 : 1. | [0.5 Mark] for P ratio 1:2. [0.5 Mark] for Q ratio 2:1. |
| Q2 | x₁ = [1(6) + 2(−3)]/3 = 0/3 = 0; y₁ = [1(−3) + 2(6)]/3 = 9/3 = 3 ⇒ P(0, 3). | [1 Mark] for evaluating P(0, 3). |
| Q3 | x₂ = [2(6) + 1(−3)]/3 = 9/3 = 3; y₂ = [2(−3) + 1(6)]/3 = 0/3 = 0 ⇒ Q(3, 0). | [1 Mark] for section formula. [1 Mark] for Q(3, 0). |
| Q3 (OR) | (−3 + x)/2 = 0 ⇒ x = 3; (6 + y)/2 = 3 ⇒ 6 + y = 6 ⇒ y = 0 ⇒ R(3, 0). | [1 Mark] for midpoint setup. [1 Mark] for R(3, 0). |
Case Study 8
Green Aviation
4 Marks
Green Aviation Waypoints
Aircraft Alpha A(−6, −4) and Aircraft Beta B(8, 6) fly along digital corridors. Waypoint W is the midpoint of AB, and weather drone P is at (2, −2).
Q1. Waypoint Location
1 Mark
Calculate the coordinates of meteorological waypoint W midway between A(−6, −4) and B(8, 6).
(A) (2, 2)
(B) (1, 1)
(C) (1, 2)
(D) (0, 1)
Q2. Drone Separation
1 Mark
Calculate direct distance between weather drone P(2, −2) and waypoint W(1, 1).
(A) √10 units (≈ 3.16)
(B) 3 units
(C) 4 units
(D) 2√3 units
Q3. Axis Division & Flight Coordinates
2 Marks
Find ratio in which y-axis divides AB and find intersection point coordinates.
(A) 4 : 3; (0, 3/7)
(B) 1 : 2; (0, 1/7)
(C) 3 : 5; (0, 2/7)
(D) 3 : 4; (0, 2/7)
OR (Alternative Q3)
If B(8, 6) is the midpoint of segment WQ where W is (1, 1), find coordinates of plane Q.
(A) (16, 12)
(B) (15, 11)
(C) (14, 10)
(D) (15, 12)
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | W = ((−6 + 8)/2, (−4 + 6)/2) = (2/2, 2/2) = (1, 1). | [1 Mark] for W(1, 1). |
| Q2 | PW = √[(1 − 2)² + (1 − (−2))²] = √[(−1)² + 3²] = √[1 + 9] = √10 units ≈ 3.16. | [1 Mark] for √10 units. |
| Q3 | 0 = [k(8) + 1(−6)]/(k + 1) ⇒ 8k = 6 ⇒ k = 3/4 (ratio 3 : 4). y = [(3/4)(6) − 4]/[(3/4) + 1] = (1/2)/(7/4) = 2/7 ⇒ (0, 2/7). |
[1 Mark] for ratio 3 : 4. [1 Mark] for (0, 2/7). |
| Q3 (OR) | (1 + x)/2 = 8 ⇒ 1 + x = 16 ⇒ x = 15; (1 + y)/2 = 6 ⇒ 1 + y = 12 ⇒ y = 11 ⇒ Q(15, 11). | [1 Mark] for midpoint equations. [1 Mark] for Q(15, 11). |
Live Practice: Chapter 7 Coordinate Geometry
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