Navigating through the newly revised CBSE Class 9 Mathematics curriculum (Ganita Manjari Part 1) requires a thorough conceptual and algebraic understanding of Coordinate Geometry, the Cartesian plane, perpendicular coordinate axes (the horizontal x-axis / abscissa and the vertical y-axis / ordinate), the origin (0, 0), the four quadrants and their respective sign conventions, point plotting, distance calculations on 2D grids, the Euclidean distance formula, midpoint determination, trisection of line segments, testing collinearity of points, verifying geometric figures (triangles, quadrilaterals, rectangles, squares), and solving real-world spatial coordinate mapping problems. Chapter 1 of Class 9 Mathematics, “Orienting Yourself: The Use of Coordinates”, introduces students to the power of algebraic geometry pioneered by French mathematician René Descartes. It transitions mathematical thinking from 1D number lines to 2D coordinate planes; demonstrates how everyday floor maps (like Reiaan’s room) can be analyzed mathematically; provides proofs for collinearity and perpendicularity; and calculates exact boundary intersections for circles and polygons. To help students master every aspect of this high-weightage chapter, this comprehensive solutions guide offers textbook-accurate, highly structured, and step-by-step responses strictly aligned with the latest CBSE Class 9 evaluation standards.
Every question presented in the official NCERT textbook—ranging from in-text “Think and Reflect” prompts and Exercise Sets 1.1 and 1.2 to the complete End-of-Chapter Exercises (Questions 1 to 16 on Pages 12–14)—has been solved with exhaustive step-by-step derivations. All numerical calculations, distance evaluations, and geometric proofs follow a structured box format with clear ASCII coordinate diagrams using clean plain-text symbols without raw LaTeX tags. Key scoring terms, official CBSE exam tags, and dynamic summary tables have been highlighted to ensure students secure maximum marks in their examinations.
Master Concept & Comparative Summary Tables
1. The Cartesian Coordinate Plane & Quadrant Sign Conventions
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THE CARTESIAN COORDINATE SYSTEM
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y-axis
▲
│ (+y)
4 ┼
QUADRANT II │ QUADRANT I
(-x, +y) │ (+x, +y)
2 ┼
│
◄───┼────┼────┼────┼────┼───────┼───────┼────┼────┼────┼────┼───► x-axis
-5 -4 -3 -2 -1 │(0,0) 1 2 3 4 5 (+x)
(-x) │ ORIGIN
-2 ┼
QUADRANT III │ QUADRANT IV
(-x, -y) │ (+x, -y)
-4 ┼
│ (-y)
▼
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| Region / Axis | Description & Condition | Sign of x-coordinate (Abscissa) | Sign of y-coordinate (Ordinate) | General Form of Point | Example Point |
|---|---|---|---|---|---|
| Quadrant I | Top-Right region (x > 0, y > 0) | Positive (+) | Positive (+) | (+x, +y) | (4, 5) |
| Quadrant II | Top-Left region (x < 0, y > 0) | Negative (-) | Positive (+) | (-x, +y) | (-3, 6) |
| Quadrant III | Bottom-Left region (x < 0, y < 0) | Negative (-) | Negative (-) | (-x, -y) | (-5, -2) |
| Quadrant IV | Bottom-Right region (x > 0, y < 0) | Positive (+) | Negative (-) | (+x, -y) | (2, -7) |
| X-axis | Horizontal reference line (y = 0) | Non-zero (x \ne 0) | Strictly Zero (0) | (x, 0) | (8, 0) |
| Y-axis | Vertical reference line (x = 0) | Strictly Zero (0) | Non-zero (y \ne 0) | (0, y) | (0, -4) |
| Origin (O) | Intersection of both axes | Zero (0) | Zero (0) | (0, 0) | (0, 0) |
2. Master Coordinate Geometry Formula Sheet
| Geometric Formula | Mathematical Expression | Key Geometric Application / Condition |
|---|---|---|
| Distance Formula | Distance d = √[ (x₂ - x₁)² + (y₂ - y₁)² ] | Calculates the straight-line distance between any two points A(x₁, y₁) and B(x₂, y₂). |
| Distance from Origin | Distance d = √(x² + y²) | Calculates the direct distance of any point P(x, y) from the origin O(0, 0). |
| Midpoint Formula | M = ( (x₁ + x₂) / 2 , (y₁ + y₂) / 2 ) | Finds the exact center point M dividing line segment AB into two equal halves (1 : 1). |
| Section Formula | P = ( (m₁·x₂ + m₂·x₁) / (m₁ + m₂) , (m₁·y₂ + m₂·y₁) / (m₁ + m₂) ) | Finds point P dividing line segment AB internally in the given ratio m₁ : m₂. |
| Trisection Points | P = ( (2·x₁ + x₂) / 3 , (2·y₁ + y₂) / 3 ) Q = ( (x₁ + 2·x₂) / 3 , (y₁ + 2·y₂) / 3 ) | Divides line segment AB into three equal segments (P divides in 1:2; Q in 2:1). |
| Collinearity Condition | AB + BC = AC (where AC is the largest length) | Proves that three distinct points A, B, and C lie along the exact same straight line. |
NCERT In-Text Questions: “Think and Reflect”
Page No. 5: Think and Reflect (Questions 1 to 3)
Question 1 What are the x-coordinate and y-coordinate of a point? How do they uniquely locate an object on a flat surface? [Exam Favorite]
Answer:
- The x-coordinate (Abscissa): The perpendicular distance of a point from the vertical y-axis, measured along or parallel to the horizontal x-axis.
- The y-coordinate (Ordinate): The perpendicular distance of a point from the horizontal x-axis, measured along or parallel to the vertical y-axis.
- Unique Location (Ordered Pair): Written together as an ordered pair (x, y), where the order of numbers is strictly fixed. The pair (x, y) represents a single, unique intersection point on the two-dimensional Cartesian plane, eliminating any spatial ambiguity.
Question 2 Is the point (3, 5) the same as the point (5, 3)? Explain with a graphical description. [Exam Favorite]
Answer: No, (3, 5) and (5, 3) represent two completely different points on the Cartesian plane.
- Point (3, 5): Located 3 units to the right of the y-axis along the x-axis, and 5 units upward parallel to the y-axis.
- Point (5, 3): Located 5 units to the right of the y-axis along the x-axis, and 3 units upward parallel to the y-axis.
- Conclusion: In coordinate geometry, the order of numbers in the pair
(x, y)matters fundamentally ((x, y) ≠ (y, x)unlessx = y).
Question 3 Can a point have a negative distance from an axis? Explain the meaning of negative coordinates. [Exam Favorite]
Answer:
- Physical Distance is Always Positive: Geometric distance is a scalar magnitude and is always positive or zero.
- Meaning of the Negative Sign: The negative sign in a coordinate indicates direction relative to the origin (0, 0):
- A negative x-coordinate (e.g.,
-4) means a distance of 4 units measured in the negative direction (to the left) of the y-axis. - A negative y-coordinate (e.g.,
-6) means a distance of 6 units measured in the downward direction below the x-axis.
- A negative x-coordinate (e.g.,
NCERT Exercise Set 1.1: Reiaan’s Room Floor Map (Page No. 7)
Question 1 Using the floor map of Reiaan’s room (where the bottom wall is aligned with the x-axis and the left wall is aligned with the y-axis): (i) If D₁R₁ represents the door to Reiaan’s room, how far is the door from the left wall (y-axis)? How far is the door from the x-axis? (ii) What are the coordinates of D₁? (iii) If R₁ is the point (11.5, 0), how wide is the door? Do you think this is a comfortable width for the room door? If a person in a wheelchair wants to enter the room, will she/he be able to do so easily? (iv) If B₁(0, 1.5) and B₂(0, 4) represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door? [Exam Favorite]
Answer:
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STEP-BY-STEP SOLUTION (EXERCISE SET 1.1):
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(i) DISTANCE OF ROOM DOOR D₁R₁ FROM AXES:
• The room door lies directly on the bottom wall (which is the x-axis).
Therefore, its perpendicular distance from the x-axis is 0 units.
• Looking at the grid, the door starts at x = 8 on the x-axis.
Therefore, the door is at a distance of 8 units from the left wall (y-axis).
(ii) COORDINATES OF POINT D₁:
• Since point D₁ is on the x-axis at a distance of 8 units to the right:
Coordinates of D₁ = (8, 0).
(iii) WIDTH OF ROOM DOOR & ACCESSIBILITY CHECK:
• Given: D₁ = (8, 0) and R₁ = (11.5, 0).
• Width of Room Door = Difference in x-coordinates = 11.5 - 8 = 3.5 units.
• Interpretation (Assuming 1 unit = 1 foot):
- 3.5 feet = 3.5 × 12 inches = 42 inches wide.
- Standard residential doors are typically 30 to 36 inches wide.
- Clear wheelchair accessibility requires a minimum doorway width of 32 inches.
- Since 42 inches > 32 inches, the door is comfortably wide and a person
in a wheelchair can enter the room easily.
(iv) COMPARISON WITH BATHROOM DOOR B₁B₂:
• Given: B₁ = (0, 1.5) and B₂ = (0, 4) on the y-axis.
• Width of Bathroom Door = Difference in y-coordinates = 4 - 1.5 = 2.5 units (30 inches).
• Comparison:
- Width of Room Door = 3.5 units (42 inches)
- Width of Bathroom Door = 2.5 units (30 inches)
- Since 2.5 < 3.5, the bathroom door is NARROWER than the room door.
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NCERT Exercise Set 1.2: Plotting Points & Shapes (Page No. 10)
Question 1 On a graph sheet, mark the x-axis and y-axis with origin O. Place Reiaan’s rectangular study table with three of its feet at the points A(8, 9), B(11, 9), and C(11, 7). (i) Where will the fourth foot of the table be? (ii) Is this placement suitable without blocking space? (iii) Calculate the width and length of the table. Can the height of the table be determined from this 2D coordinate map? [Exam Favorite]
Answer:
================================================================================ STEP-BY-STEP SOLUTION (EXERCISE SET 1.2 - QUESTION 1): -------------------------------------------------------------------------------- (i) FINDING THE FOURTH VERTEX D: • In a rectangle ABCD, opposite sides are parallel and equal in length. • Points A(8, 9) and B(11, 9) lie on the horizontal line y = 9. Horizontal length AB = 11 - 8 = 3 units. • Points B(11, 9) and C(11, 7) lie on the vertical line x = 11. Vertical width BC = 9 - 7 = 2 units. • The fourth vertex D must share: - The same x-coordinate as point A: x = 8. - The same y-coordinate as point C: y = 7. • Coordinates of fourth foot D = (8, 7). (ii) SPATIAL SUITABILITY: • Yes, the placement is suitable as it aligns neatly in the upper corner of the room without obstructing the door opening D₁R₁(8 to 11.5 along y = 0). (iii) DIMENSIONS OF STUDY TABLE: • Width (Horizontal dimension) = |11 - 8| = 3 units. • Length (Vertical dimension) = |9 - 7| = 2 units. • Area of table top = 3 × 2 = 6 square units. • Height of the Table: CANNOT BE DETERMINED. A 2D floor plan only provides length (x) and width (y), giving zero information about vertical height (z-axis). ================================================================================
NCERT End-of-Chapter Exercises (Pages 12–14, Questions 1 to 16)
Question 1 What are the x-coordinate and y-coordinate of the point of intersection of the two axes? [Exam Favorite]
Answer:
- The horizontal x-axis and vertical y-axis intersect at the Origin (O).
- x-coordinate = 0 and y-coordinate = 0.
- The coordinates of the point of intersection are (0, 0).
Question 2 Point W has x-coordinate equal to -5. Can you predict the coordinates of point H which is on the line through W parallel to the y-axis? Which quadrants can H lie in? [Exam Favorite]
Answer:
- Coordinates of Point H: A line parallel to the y-axis is vertical; every point on this line has a constant x-coordinate equal to
-5. Therefore, point H has coordinates of the form (-5, y), where y is any real number. - Possible Quadrants for H:
- If y > 0 (positive), point H(-5, y) lies in Quadrant II.
- If y < 0 (negative), point H(-5, y) lies in Quadrant III.
- If y = 0, point H(-5, 0) lies directly on the Negative X-axis.
Question 3 Consider the points R(3, 0), A(0, -2), M(-5, -2), and P(-5, 2). If they are joined in the same order to form quadrilateral RAMP, predict: (i) Two sides of RAMP that are perpendicular to each other. (ii) A side of RAMP that is parallel to an axis. (iii) Two vertices that lie on a line parallel to an axis. [Exam Favorite]
Answer:
================================================================================ STEP-BY-STEP GEOMETRIC ANALYSIS OF QUADRILATERAL RAMP: -------------------------------------------------------------------------------- GIVEN VERTICES: R(3, 0), A(0, -2), M(-5, -2), P(-5, 2) (i) TWO SIDES PERPENDICULAR TO EACH OTHER: • Side AM connects A(0, -2) and M(-5, -2): Both y-coordinates are -2 (Horizontal line y = -2). • Side MP connects M(-5, -2) and P(-5, 2): Both x-coordinates are -5 (Vertical line x = -5). • Since a horizontal line and a vertical line meet at 90 degrees: Side AM is strictly PERPENDICULAR to Side MP (AM ⟂ MP). (ii) SIDE PARALLEL TO AN AXIS: • Side AM lies on y = -2, which is PARALLEL to the x-axis. • Side MP lies on x = -5, which is PARALLEL to the y-axis. (iii) TWO VERTICES ON A LINE PARALLEL TO AN AXIS: • Vertices M(-5, -2) and P(-5, 2) lie on the line x = -5 (Parallel to y-axis). • Vertices A(0, -2) and M(-5, -2) lie on the line y = -2 (Parallel to x-axis). ================================================================================
Question 4 Plot point Z(5, -6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides. [Exam Favorite]
Answer:
================================================================================ STEP-BY-STEP RIGHT-ANGLED TRIANGLE IZN CONSTRUCTION: -------------------------------------------------------------------------------- CONSTRUCTION: • Let Point Z = (5, -6) • Drop a perpendicular from Z to the x-axis to get Point I = (5, 0). • Drop a perpendicular from Z to the y-axis to get Point N = (0, -6). • Join I(5, 0), Z(5, -6), and N(0, -6) to form right-angled triangle IZN. LENGTHS OF THE THREE SIDES: 1. Side IZ (Vertical leg): IZ = |0 - (-6)| = 6 units. 2. Side ZN (Horizontal leg): ZN = |5 - 0| = 5 units. 3. Side IN (Hypotenuse): Using Pythagoras Theorem: IN² = IZ² + ZN² IN² = (6)² + (5)² = 36 + 25 = 61 IN = √61 units ≈ 7.81 units. FINAL ANSWER: • Side IZ = 6 units • Side ZN = 5 units • Side IN = √61 units ================================================================================
Question 5 What would a system of coordinates be like if we did not have negative numbers? Would this system allow us to locate all the points on a 2-D plane? [Exam Favorite]
Answer:
- Nature of the System: Without negative numbers, both the x-coordinate and y-coordinate would be restricted strictly to non-negative values (x \ge 0, y \ge 0).
- Can it locate all points on a 2D plane? NO.
- Explanation: A system with only positive coordinates can represent only Quadrant I. Points situated to the left of the y-axis (Quadrants II and III) or below the x-axis (Quadrants III and IV) could never be described or located. Negative numbers are essential to provide complete 360-degree spatial coverage in two dimensions.
Question 6 Are the points M(-3, -4), A(0, 0), and G(6, 8) on the same straight line (collinear)? Suggest a method to check this without plotting and joining the points. [Exam Favorite]
Answer:
================================================================================ COLLINEARITY PROOF USING DISTANCE METHOD: -------------------------------------------------------------------------------- GIVEN POINTS: M(-3, -4), A(0, 0), G(6, 8) Distance Formula: d = √[ (x₂ - x₁)² + (y₂ - y₁)² ] STEP 1: Calculate Distance MA MA = √[ (0 - (-3))² + (0 - (-4))² ] MA = √[ (3)² + (4)² ] = √(9 + 16) = √25 = 5 units STEP 2: Calculate Distance AG AG = √[ (6 - 0)² + (8 - 0)² ] AG = √[ (6)² + (8)² ] = √(36 + 64) = √100 = 10 units STEP 3: Calculate Distance MG MG = √[ (6 - (-3))² + (8 - (-4))² ] MG = √[ (9)² + (12)² ] = √(81 + 144) = √225 = 15 units STEP 4: Check Collinearity Condition Sum of two smaller segments = MA + AG = 5 + 10 = 15 units. Longest segment = MG = 15 units. Since MA + AG = MG (15 = 15): CONCLUSION: The points M(-3, -4), A(0, 0), and G(6, 8) are strictly COLLINEAR (lie on the same straight line). ================================================================================
Question 7 Use your method (from Problem 6) to check if the points R(-5, -1), B(-2, -5), and C(4, -12) are on the same straight line. [Exam Favorite]
Answer:
================================================================================ COLLINEARITY CHECK (POINTS R, B, C): -------------------------------------------------------------------------------- GIVEN POINTS: R(-5, -1), B(-2, -5), C(4, -12) STEP 1: Calculate Distance RB RB = √[ (-2 - (-5))² + (-5 - (-1))² ] RB = √[ (3)² + (-4)² ] = √(9 + 16) = √25 = 5 units STEP 2: Calculate Distance BC BC = √[ (4 - (-2))² + (-12 - (-5))² ] BC = √[ (6)² + (-7)² ] = √(36 + 49) = √85 ≈ 9.22 units STEP 3: Calculate Distance RC RC = √[ (4 - (-5))² + (-12 - (-1))² ] RC = √[ (9)² + (-11)² ] = √(81 + 121) = √202 ≈ 14.21 units STEP 4: Compare Sums RB + BC = 5 + √85 ≈ 5 + 9.22 = 14.22 units. RC = √202 ≈ 14.2127 units. Notice that (5 + √85)² = 25 + 85 + 10√85 = 110 + 92.195 = 202.195 ≠ 202. Since RB + BC ≠ RC: CONCLUSION: The points R(-5, -1), B(-2, -5), and C(4, -12) do NOT lie on the same straight line (they are NOT collinear). ================================================================================
Question 8 Using the origin O(0, 0) as one vertex, plot the vertices of: (i) A right-angled isosceles triangle in Quadrant I. (ii) An isosceles triangle whose base lies along the negative y-axis. [Exam Favorite]
Answer:
- (i) Right-Angled Isosceles Triangle in Quadrant I:
- Vertex 1:
O(0, 0)at the origin. - Vertex 2:
A(4, 0)on the positive x-axis (Length OA = 4 units). - Vertex 3:
B(0, 4)on the positive y-axis (Length OB = 4 units). - Triangle OAB has \angle AOB = 90^\circ and equal legs OA = OB = 4 units.
- Vertex 1:
- (ii) Isosceles Triangle with Base along Negative Y-axis:
- Vertex 1:
P(-4, -3)in Quadrant III. - Vertex 2:
Q(0, -1)on the negative y-axis. - Vertex 3:
R(0, -5)on the negative y-axis. - Base QR lies on the y-axis with length
|-1 - (-5)| = 4units, and apex P is equidistant from Q and R.
- Vertex 1:
Question 9 Check whether point M(2, 3) is the exact midpoint of line segment ST with endpoints S(-2, 1) and T(6, 5). [Exam Favorite]
Answer:
================================================================================ MIDPOINT VERIFICATION: -------------------------------------------------------------------------------- GIVEN ENDPOINTS: S(-2, 1) and T(6, 5) Midpoint Formula: M(x, y) = ( (x₁ + x₂) / 2 , (y₁ + y₂) / 2 ) x_mid = (-2 + 6) / 2 = 4 / 2 = 2 y_mid = (1 + 5) / 2 = 6 / 2 = 3 Calculated Midpoint = (2, 3) Given Point M = (2, 3) CONCLUSION: YES, Point M(2, 3) is strictly the exact midpoint of line segment ST. ================================================================================
Question 10 Point M(-7, 1) is the midpoint of line segment AB. If the coordinates of endpoint A are (3, -4), find the coordinates of endpoint B. [Exam Favorite]
Answer:
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FINDING UNKNOWN ENDPOINT USING MIDPOINT FORMULA:
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GIVEN DATA:
• Endpoint A = (3, -4) ===> x₁ = 3, y₁ = -4
• Midpoint M = (-7, 1) ===> x_m = -7, y_m = 1
• Let Endpoint B = (x, y)
STEP 1: Solving for x-coordinate
Formula: x_m = (x₁ + x) / 2
-7 = (3 + x) / 2
Multiply by 2:
-14 = 3 + x
x = -14 - 3 = -17
STEP 2: Solving for y-coordinate
Formula: y_m = (y₁ + y) / 2
1 = (-4 + y) / 2
Multiply by 2:
2 = -4 + y
y = 2 + 4 = 6
FINAL ANSWER:
The coordinates of endpoint B are (-17, 6).
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Question 11 Find the coordinates of the points of trisection (points P and Q dividing in three equal parts) of the line segment joining A(4, 7) and B(16, -2). [Exam Favorite]
Answer:
================================================================================ TRISECTION OF LINE SEGMENT AB: -------------------------------------------------------------------------------- GIVEN ENDPOINTS: A(4, 7) and B(16, -2) Points P and Q divide segment AB into three equal parts (AP = PQ = QB). • Point P divides AB internally in the ratio 1 : 2. • Point Q divides AB internally in the ratio 2 : 1. STEP 1: Finding Coordinates of Point P (Ratio 1 : 2) Section Formula: P = ( (1·x₂ + 2·x₁) / (1 + 2) , (1·y₂ + 2·y₁) / (1 + 2) ) x_p = [ (1 × 16) + (2 × 4) ] / 3 = (16 + 8) / 3 = 24 / 3 = 8 y_p = [ (1 × (-2)) + (2 × 7) ] / 3 = (-2 + 14) / 3 = 12 / 3 = 4 Coordinates of Point P = (8, 4). STEP 2: Finding Coordinates of Point Q (Ratio 2 : 1) Section Formula: Q = ( (2·x₂ + 1·x₁) / (2 + 1) , (2·y₂ + 1·y₁) / (2 + 1) ) x_q = [ (2 × 16) + (1 × 4) ] / 3 = (32 + 4) / 3 = 36 / 3 = 12 y_q = [ (2 × (-2)) + (1 × 7) ] / 3 = (-4 + 7) / 3 = 3 / 3 = 1 Coordinates of Point Q = (12, 1). FINAL ANSWER: The points of trisection are P(8, 4) and Q(12, 1). ================================================================================
Question 12 A circle with center at origin O(0, 0) passes through point A(7, 4). (i) Find the radius of the circle. (ii) Determine whether point D(2, 6) and point E(0, 9) lie inside, on, or outside the circle. [Exam Favorite]
Answer:
================================================================================ CIRCLE RADIUS & POINT POSITION VERIFICATION: -------------------------------------------------------------------------------- (i) RADIUS OF THE CIRCLE: Radius r = Distance OA = √[ (7 - 0)² + (4 - 0)² ] r = √(7² + 4²) = √(49 + 16) = √65 units (r² = 65 ≈ 8.06 units). (ii) POSITION OF POINTS D(2, 6) AND E(0, 9): • For Point D(2, 6): Distance OD = √(2² + 6²) = √(4 + 36) = √40 Since OD (√40) < Radius (√65): Point D(2, 6) lies INSIDE the circle. • For Point E(0, 9): Distance OE = √(0² + 9²) = √81 = 9 units Since OE (√81) > Radius (√65): Point E(0, 9) lies OUTSIDE the circle. FINAL ANSWER: • Radius of circle = √65 units • Point D(2, 6) lies INSIDE the circle • Point E(0, 9) lies OUTSIDE the circle ================================================================================
Question 13 Find the coordinates of the vertices of a triangle ABC, given that the midpoints of its sides BC, CA, and AB are D(5, 1), E(6, 5), and F(0, 3) respectively. [Exam Favorite]
Answer:
================================================================================ RECOVERING TRIANGLE VERTICES FROM MIDPOINTS: -------------------------------------------------------------------------------- Let the vertices of triangle ABC be A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃). GIVEN MIDPOINTS: • Midpoint of BC is D(5, 1) ===> x₂ + x₃ = 10 ...(1), y₂ + y₃ = 2 ...(2) • Midpoint of CA is E(6, 5) ===> x₃ + x₁ = 12 ...(3), y₃ + y₁ = 10 ...(4) • Midpoint of AB is F(0, 3) ===> x₁ + x₂ = 0 ...(5), y₁ + y₂ = 6 ...(6) STEP 1: Solving for x-coordinates Add equations (1), (3), and (5): 2·(x₁ + x₂ + x₃) = 10 + 12 + 0 = 22 x₁ + x₂ + x₃ = 11 ...(7) • x₁ = (x₁ + x₂ + x₃) - (x₂ + x₃) = 11 - 10 = 1 • x₂ = (x₁ + x₂ + x₃) - (x₃ + x₁) = 11 - 12 = -1 • x₃ = (x₁ + x₂ + x₃) - (x₁ + x₂) = 11 - 0 = 11 STEP 2: Solving for y-coordinates Add equations (2), (4), and (6): 2·(y₁ + y₂ + y₃) = 2 + 10 + 6 = 18 y₁ + y₂ + y₃ = 9 ...(8) • y₁ = (y₁ + y₂ + y₃) - (y₂ + y₃) = 9 - 2 = 7 • y₂ = (y₁ + y₂ + y₃) - (y₃ + y₁) = 9 - 10 = -1 • y₃ = (y₁ + y₂ + y₃) - (y₁ + y₂) = 9 - 6 = 3 FINAL ANSWER: The vertices of triangle ABC are A(1, 7), B(-1, -1), and C(11, 3). ================================================================================
Question 14 In a city grid layout where streets run along coordinate lines, each intersection is marked by coordinates (x, y). How many distinct street routes of shortest path length exist between intersection (1, 1) and intersection (3, 2)? [Exam Favorite]
Answer:
================================================================================ GRID PATH CALCULATION (MANHATTAN DISTANCE): -------------------------------------------------------------------------------- From Start (1, 1) to Destination (3, 2): • Horizontal blocks to move east = 3 - 1 = 2 steps (E, E) • Vertical blocks to move north = 2 - 1 = 1 step (N) • Total steps required = 2 + 1 = 3 steps Total unique shortest paths = Combinations of (E, E, N) Path 1: East -> East -> North (E, E, N) Path 2: East -> North -> East (E, N, E) Path 3: North -> East -> East (N, E, E) Formula: Total Paths = 3! / (2! × 1!) = 6 / 2 = 3 routes. FINAL ANSWER: There are 3 distinct shortest street routes between (1, 1) and (3, 2). ================================================================================
Question 15 Two radar circles are centered at A(100, 150) with radius r₁ = 80 units and B(250, 230) with radius r₂ = 100 units on an 800 × 600 display screen. (i) Verify whether any part of either circle lies outside the screen display. (ii) Determine whether the two radar circles intersect each other. [Exam Favorite]
Answer:
================================================================================ RADAR CIRCLE INTERSECTION & SCREEN BOUNDARY ANALYSIS: -------------------------------------------------------------------------------- (i) SCREEN BOUNDARY CHECK (Screen bounds: 0 ≤ x ≤ 800, 0 ≤ y ≤ 600): • Circle A (Center 100, 150; r₁ = 80): - x-range: [100 - 80, 100 + 80] = [20, 180] (Within 0 to 800) - y-range: [150 - 80, 150 + 80] = [70, 230] (Within 0 to 600) ===> Circle A is FULLY inside the screen. • Circle B (Center 250, 230; r₂ = 100): - x-range: [250 - 100, 250 + 100] = [150, 350] (Within 0 to 800) - y-range: [230 - 100, 230 + 100] = [130, 330] (Within 0 to 600) ===> Circle B is FULLY inside the screen. (ii) CIRCLE INTERSECTION TEST: • Distance between centers d(AB): d = √[ (250 - 100)² + (230 - 150)² ] d = √(150² + 80²) = √(22500 + 6400) = √28900 = 170 units. • Sum of Radii = r₁ + r₂ = 80 + 100 = 180 units. • Difference of Radii = |r₁ - r₂| = |80 - 100| = 20 units. • Intersection Condition: Since |r₁ - r₂| < d < (r₁ + r₂) ===> 20 < 170 < 180. FINAL ANSWER: (i) Neither circle lies outside the screen boundary. (ii) The two radar circles INTERSECT each other at two distinct points. ================================================================================
Question 16 Show that the points A(2, 1), B(-1, 2), C(-2, -1), and D(1, -2) are the vertices of a square. Also, find the area of the square ABCD. [Exam Favorite]
Answer:
================================================================================ SQUARE VERIFICATION USING DISTANCE FORMULA: -------------------------------------------------------------------------------- GIVEN VERTICES: A(2, 1), B(-1, 2), C(-2, -1), D(1, -2) STEP 1: Calculate Lengths of all Four Sides • AB = √[ (-1 - 2)² + (2 - 1)² ] = √[ (-3)² + (1)² ] = √(9 + 1) = √10 units • BC = √[ (-2 - (-1))² + (-1 - 2)² ] = √[ (-1)² + (-3)² ] = √(1 + 9) = √10 units • CD = √[ (1 - (-2))² + (-2 - (-1))² ] = √[ (3)² + (-1)² ] = √(9 + 1) = √10 units • DA = √[ (2 - 1)² + (1 - (-2))² ] = √[ (1)² + (3)² ] = √(1 + 9) = √10 units ===> All four sides are strictly equal: AB = BC = CD = DA = √10 units. STEP 2: Calculate Lengths of both Diagonals • Diagonal AC = √[ (-2 - 2)² + (-1 - 1)² ] = √[ (-4)² + (-2)² ] = √(16 + 4) = √20 units • Diagonal BD = √[ (1 - (-1))² + (-2 - 2)² ] = √[ (2)² + (-4)² ] = √(4 + 16) = √20 units ===> Both diagonals are strictly equal: Diagonal AC = Diagonal BD = √20 units. CONCLUSION: Since all 4 sides are equal (√10) and both diagonals are equal (√20), ABCD is strictly a SQUARE. STEP 3: Calculate Area of Square ABCD Area = (Side)² = (√10)² = 10 square units. FINAL ANSWER: • ABCD is verified to be a Square. • Area of Square ABCD = 10 square units. ================================================================================
Frequently Asked Questions (FAQs) – Class 9 Maths Chapter 1
Question 1: Who is credited with inventing Coordinate Geometry? [Exam Favorite] Answer: French mathematician and philosopher René Descartes (1596–1650) developed Coordinate Geometry, which is why the system is called the Cartesian Coordinate System.
Question 2: What is the Abscissa and Ordinate of the point (-7, 4)? [Exam Favorite] Answer:
- Abscissa (x-coordinate): -7
- Ordinate (y-coordinate): 4
Question 3: In which quadrant does the point (-4, -9) lie? [Exam Favorite] Answer: Since both coordinates are negative (x < 0, y < 0), the point lies in Quadrant III.
Question 4: What are the coordinates of a point lying on the y-axis at a distance of 6 units below the origin? [Exam Favorite] Answer: On the y-axis, the x-coordinate is always zero. Below the origin means y = -6. The coordinates are (0, -6).
Question 5: What is the distance of the point P(5, -12) from the origin? [Exam Favorite] Answer: Distance = √(x² + y²) = √(5² + (-12)²) = √(25 + 144) = √169 = 13 units.
Question 6: If the coordinates of two points are A(3, 0) and B(-5, 0), what is the length of AB? [Exam Favorite] Answer: Length AB = |x₂ - x₁| = |-5 - 3| = |-8| = 8 units.
Question 7: What is the perpendicular distance of the point P(-6, 8) from the x-axis and y-axis? [Exam Favorite] Answer:
- Perpendicular distance from x-axis =
|y-coordinate| = |8| =8 units. - Perpendicular distance from y-axis =
|x-coordinate| = |-6| =6 units.
Question 8: Write the coordinates of the mirror reflection of the point (3, -5) across: (a) X-axis, (b) Y-axis. [Exam Favorite] Answer:
- (a) Reflection in X-axis (invert sign of y): (3, 5).
- (b) Reflection in Y-axis (invert sign of x): (-3, -5).
Question 9: If the coordinates of three vertices of rectangle OABC are O(0, 0), A(6, 0), and C(0, 4), find vertex B. [Exam Favorite] Answer: Vertex B shares the x-coordinate of A and the y-coordinate of C. Therefore, B = (6, 4).
Question 10: Find the midpoint of the line segment joining (8, -2) and (-4, 6). [Exam Favorite] Answer: Midpoint = ( (8 + (-4)) / 2 , (-2 + 6) / 2 ) = ( 4 / 2 , 4 / 2 ) = (2, 2).
Question 11: Can the distance between two points ever be negative? [Exam Favorite] Answer: No. Distance is a geometric length and is always non-negative (\ge 0).
Question 12: What is the equation of the x-axis and y-axis? [Exam Favorite] Answer:
- Equation of the X-axis: y = 0
- Equation of the Y-axis: x = 0
Question 13: Under what condition do three points A, B, and C form a triangle? [Exam Favorite] Answer: Three points form a triangle when they are non-collinear (AB + BC > AC).
Question 14: What is the ordinate of all points lying on the horizontal line passing through (2, 7)? [Exam Favorite] Answer: Every point on this horizontal line has an ordinate equal to 7 (y = 7).
Question 15: If the area of a square formed by vertices on a coordinate grid is 25 square units, what is the side length? [Exam Favorite] Answer: Side = √(Area) = √25 = 5 units.
Mastering the NCERT Solutions for Class 9 Mathematics Chapter 1 (Ganita Manjari Part 1), “Orienting Yourself: The Use of Coordinates”, equips students with the Cartesian plane concepts, distance and midpoint formulas, geometric proofs, and spatial coordinate problem-solving techniques required for top performance in CBSE examinations. Review the step-by-step box solutions, the master summary tables, and the 15 high-yield FAQs above to secure full marks in your examinations.