CBSE Class 9 Maths Ganita Manjari Part 1 Ch 01 Orienting Yourself The Use of Coordinates Long Answer and Case Study Questions with Answers

5 Mark Questions

Question 1. The midpoints of the sides of triangle ABC are D(5, 1), E(6, 5) and F(0, 3).
(i) Let the unknown vertices be A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃). Set up a system of linear equations relating the vertex coordinates to the given midpoints D, E and F.
(ii) Solve the system step-by-step to determine the exact coordinates of vertices A, B and C.
(iii) Verify your result by calculating the midpoint of segment AB and matching it with the given midpoint.
Answer:
(i) Take D as the midpoint of BC, E as the midpoint of AC and F as the midpoint of AB.
D: \( \frac{x_2 + x_3}{2} = 5 \Rightarrow x_2 + x_3 = 10 \) … (1) and \( y_2 + y_3 = 2 \) … (2)
E: \( \frac{x_1 + x_3}{2} = 6 \Rightarrow x_1 + x_3 = 12 \) … (3) and \( y_1 + y_3 = 10 \) … (4)
F: \( \frac{x_1 + x_2}{2} = 0 \Rightarrow x_1 + x_2 = 0 \) … (5) and \( y_1 + y_2 = 6 \) … (6)
(ii) Adding (1), (3) and (5): \( 2(x_1 + x_2 + x_3) = 22 \Rightarrow x_1 + x_2 + x_3 = 11 \) … (7)
From (7) − (1): \( x_1 = 1 \). From (7) − (3): \( x_2 = -1 \). From (7) − (5): \( x_3 = 11 \).
Adding (2), (4) and (6): \( 2(y_1 + y_2 + y_3) = 18 \Rightarrow y_1 + y_2 + y_3 = 9 \) … (8)
From (8) − (2): \( y_1 = 7 \). From (8) − (4): \( y_2 = -1 \). From (8) − (6): \( y_3 = 3 \).
\( \therefore A(1, 7),\ B(-1, -1),\ C(11, 3) \)
(iii) Midpoint of AB = \( \left( \frac{1 + (-1)}{2}, \frac{7 + (-1)}{2} \right) = (0, 3) \), which is F. ✓
Extra check: the midpoint of BC is (5, 1) = D ✓, and the midpoint of AC is (6, 5) = E ✓.
In simple words: Each midpoint gives two equations. Adding all three x-equations gives the total of the x-values, and subtracting each equation from that total leaves one vertex at a time.

Exam Tip: Clearly state which midpoint belongs to which side before writing equations. A mismatch here changes every answer.

Question 2. A city layout is designed as a square grid centred at origin O(0, 0). Two main perpendicular avenues run along the axes: East-West Avenue (x-axis) and North-South Avenue (y-axis). Parallel streets are built at uniform intervals of 200 metres (1 unit on grid = 200 m).
A cyclist starts at intersection C(−4, 3) and travels along grid lines to D(5, −9).
(i) Calculate the Manhattan distance (distance travelled strictly along grid lines) in metres.
(ii) Calculate the Euclidean (direct shortest crow-fly) distance in metres between C and D.
(iii) Explain how this urban grid system reflects the historical urban planning of the ancient Sindhu-Sarasvatī Civilisation.
Answer:
(i) Horizontal movement = \( |5 - (-4)| = 9 \) units. Vertical movement = \( |-9 - 3| = 12 \) units.
Manhattan distance = 9 + 12 = 21 units = \( 21 \times 200 \) = 4,200 m (4.2 km)
(ii) Euclidean distance = \( \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15 \) units = \( 15 \times 200 \) = 3,000 m (3 km)
(iii) Cities of the Sindhu-Sarasvatī (Harappan) Civilisation, such as Mohenjo-daro, were planned with main streets running roughly north–south and east–west and crossing at right angles. This divided the city into rectangular blocks. Just like our grid, a location could be described by how many blocks east–west and north–south it lay from a reference point. So the idea of two perpendicular reference directions, which is the basis of the coordinate plane, was already used in their town planning.
In simple words: Along the streets the cyclist rides 4.2 km, but flying straight it is only 3 km. Harappan cities were laid out in the same criss-cross grid pattern.

Exam Tip: Convert grid units to metres only at the end. Multiply the final unit answer by 200 to avoid errors.

Question 3. Consider circle K centred at the origin O(0, 0) passing through point A(1, −8).
(i) Prove that points B(−4, 7) and C(−7, −4) also lie on circle K.
(ii) Check whether point D(−5, 6) lies inside, on, or outside circle K.
(iii) Check whether point E(0, 9) lies inside, on, or outside circle K.
(iv) Calculate the exact area of the circle K in square units (leave answer in terms of π).
Answer:
Radius: \( r^2 = OA^2 = 1^2 + (-8)^2 = 65 \), so \( r = \sqrt{65} \).
(i) \( OB^2 = (-4)^2 + 7^2 = 16 + 49 = 65 = r^2 \), so B lies on K.
\( OC^2 = (-7)^2 + (-4)^2 = 49 + 16 = 65 = r^2 \), so C lies on K.
(ii) \( OD^2 = (-5)^2 + 6^2 = 25 + 36 = 61 \). Since 61 < 65, D lies inside circle K.
(iii) \( OE^2 = 0^2 + 9^2 = 81 \). Since 81 > 65, E lies outside circle K.
(iv) Area = \( \pi r^2 = 65\pi \) square units (about 204.2 sq units).
In simple words: Compare each point's squared distance from the centre with 65. Equal means on the circle, smaller means inside, larger means outside.

Exam Tip: Comparing squared distances with \( r^2 \) avoids working with square roots and is just as valid.

Question 4. Given quadrilateral ABCD with vertices A(2, 1), B(−1, 2), C(−2, −1) and D(1, −2), use the Baudhāyana-Pythagoras distance formula to:
(i) Calculate the lengths of all four sides AB, BC, CD and DA.
(ii) Calculate the lengths of both diagonals AC and BD.
(iii) Justify whether ABCD is a square.
(iv) Compute the exact area of quadrilateral ABCD.
Answer:
(i) \( AB = \sqrt{(-1-2)^2 + (2-1)^2} = \sqrt{9 + 1} = \sqrt{10} \)
\( BC = \sqrt{(-2+1)^2 + (-1-2)^2} = \sqrt{1 + 9} = \sqrt{10} \)
\( CD = \sqrt{(1+2)^2 + (-2+1)^2} = \sqrt{9 + 1} = \sqrt{10} \)
\( DA = \sqrt{(2-1)^2 + (1+2)^2} = \sqrt{1 + 9} = \sqrt{10} \)
(ii) \( AC = \sqrt{(-2-2)^2 + (-1-1)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \)
\( BD = \sqrt{(1+1)^2 + (-2-2)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \)
(iii) All four sides are equal, so ABCD is at least a rhombus. Its diagonals are also equal, and a rhombus with equal diagonals is a square. \( \therefore \) ABCD is a square.
(Check: \( AB^2 + BC^2 = 10 + 10 = 20 = AC^2 \), so ∠B = 90°.)
(iv) Area = side² = \( (\sqrt{10})^2 \) = 10 square units
In simple words: Equal sides and equal diagonals together prove a square, and its area is the side multiplied by itself.

Exam Tip: Equal sides alone only prove a rhombus. You must also show equal diagonals, or one right angle, to prove a square.

Case Study Based Questions

Question 5. Reiaan and Shalini's family recently moved into a new home. Shalini, having studied Coordinate Geometry in Grade 9, designed a tactile grid floor map of Reiaan's bedroom and bathroom to help him navigate independently. The floor plan is modelled on a 2D Cartesian plane, where 1 grid unit = 1 foot.
Bedroom vertices: O(0, 0), A(12, 0), B(12, 10), C(0, 10)
Main entrance door spans segment D₁R₁ along the x-axis: D₁(8, 0), R₁(11.5, 0)
Bathroom door spans segment B₁B₂ along the y-axis: B₁(0, 1.5), B₂(0, 4)
Wardrobe occupies space between: W₁(3, 0), W₂(7, 0), W₄(3, 2)
(i) Calculate the width of the main entrance door D₁R₁ and the bathroom door B₁B₂.
(ii) Standard wheelchair accessibility guidelines require interior doors to have a minimum clearance width of 3.0 ft. Evaluate whether both doors meet accessibility standards for a person using a wheelchair.
(iii) If the bathroom door hinges at B₁(0, 1.5) and swings 90° outward into the bedroom along a circular arc of radius equal to the door width, determine the coordinates of the door's outer edge when fully opened into the room. Check if it collides with the wardrobe located between W₁(3, 0) and W₂(7, 0) extending up to W₄(3, 2).
Answer:
(i) Entrance door D₁R₁ lies on the x-axis: width = \( |11.5 - 8| \) = 3.5 ft
Bathroom door B₁B₂ lies on the y-axis: width = \( |4 - 1.5| \) = 2.5 ft
(ii) Entrance door: 3.5 ft ≥ 3.0 ft, so it meets the standard.
Bathroom door: 2.5 ft < 3.0 ft, so it does not meet the standard. It is 0.5 ft too narrow for a wheelchair.
(iii) The closed door lies along the y-axis from B₁(0, 1.5) up to its outer edge B₂(0, 4). Turning it 90° about the hinge into the bedroom (towards positive x) swings the outer edge down to the same height as the hinge, 2.5 ft to the right.
Outer edge when fully open = (0 + 2.5, 1.5) = (2.5, 1.5)
While swinging, every point of the door stays within 2.5 ft of the hinge, so the door never goes beyond x = 2.5.
The wardrobe starts at x = 3, and 2.5 < 3.
\( \therefore \) The door does not collide with the wardrobe. It stops 0.5 ft short of it.
In simple words: The entrance door is wide enough for a wheelchair but the bathroom door is not. When the bathroom door opens fully, its edge reaches x = 2.5, just short of the wardrobe at x = 3.

Exam Tip: For a door on the y-axis, width is the difference of the y-coordinates. For a door on the x-axis, it is the difference of the x-coordinates.

Question 6. Throughout history, coordinate systems developed across cultures for sky mapping and terrestrial navigation. In 5th-century India, Āryabhaṭa used Celestial Coordinates based on the ecliptic and introduced sine functions. Later, Islamic scholar Al-Bīrūnī (c. 1000 CE) perfected the handheld 'astrolabe' to calculate city coordinates using Indian trigonometric methods. Modern navigation systems rely on similar Cartesian coordinates overlaying geographical zones.
Suppose a regional navigation grid places a coastal port at origin O(0, 0). Lighthouse L₁ is at (4, 3), Lighthouse L₂ is at (−5, 12), and a cargo vessel V is at (12, −5).
(i) Calculate the direct distance of Lighthouse L₁ and Lighthouse L₂ from the port origin O(0, 0).
(ii) In which quadrants do Lighthouse L₂ and Vessel V lie?
(iii) The cargo vessel V(12, −5) sends a distress signal. Determine which lighthouse (L₁ or L₂) is closer to vessel V in direct line-of-sight distance. Support your answer with full calculations.
Answer:
(i) \( OL_1 = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} \) = 5 units
\( OL_2 = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} \) = 13 units
(ii) L₂(−5, 12): x < 0 and y > 0, so it is in Quadrant II.
V(12, −5): x > 0 and y < 0, so it is in Quadrant IV.
(iii) \( VL_1 = \sqrt{(12-4)^2 + (-5-3)^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2} \approx 11.31 \) units
\( VL_2 = \sqrt{(12+5)^2 + (-5-12)^2} = \sqrt{289 + 289} = \sqrt{578} = 17\sqrt{2} \approx 24.04 \) units
Since 11.31 < 24.04, Lighthouse L₁ is closer to the vessel, by about 12.73 units.
In simple words: L₁ is less than half as far from the ship as L₂, so L₁ should respond to the distress signal.

Exam Tip: Subtracting a negative number is the most common slip here. For example, 12 − (−5) = 17, not 7.

Question 7. A computer graphics application renders a graphical user interface (GUI) on a display screen of size 800 × 600 pixels. The origin (0, 0) is located at the bottom-left corner of the screen. x increases horizontally to the right (up to 800), and y increases vertically upward (up to 600).
Two circular interactive icons are rendered on the screen:
• Icon A (Settings): Centre at A(100, 150), Radius R₁ = 80 pixels.
• Icon B (Profile): Centre at B(250, 230), Radius R₂ = 100 pixels.
(i) Determine whether any part of Icon A or Icon B extends outside the visible boundary of the screen \( [0 \le x \le 800,\ 0 \le y \le 600] \). Show analytical proof.
(ii) Calculate the straight-line distance between the centres A and B of the two icons.
(iii) Determine whether Icon A and Icon B overlap each other on the screen.
(iv) If a software engineer wants to move Icon B horizontally to the right along y = 230 so that it is exactly tangent to Icon A (touching externally at one point), what should be the new x-coordinate of Icon B's centre?
Answer:
(i) Icon A covers x from 100 − 80 = 20 to 100 + 80 = 180, and y from 150 − 80 = 70 to 150 + 80 = 230.
Icon B covers x from 250 − 100 = 150 to 250 + 100 = 350, and y from 230 − 100 = 130 to 230 + 100 = 330.
All these values lie between 0 and 800 for x and between 0 and 600 for y. \( \therefore \) Neither icon extends outside the screen.
(ii) \( AB = \sqrt{(250-100)^2 + (230-150)^2} = \sqrt{22500 + 6400} = \sqrt{28900} \) = 170 pixels
(iii) R₁ + R₂ = 180 pixels. Since AB = 170 < 180, the icons overlap.
(iv) For external tangency, the new distance between centres must equal R₁ + R₂ = 180. Let the new centre be (x, 230).
\( (x - 100)^2 + (230 - 150)^2 = 180^2 \)
\( (x - 100)^2 + 6400 = 32400 \)
\( (x - 100)^2 = 26000 \Rightarrow x - 100 = \sqrt{26000} = 20\sqrt{65} \approx 161.25 \) (taking the positive root, since B moves to the right)
\( \therefore x = 100 + 20\sqrt{65} \approx 261.25 \) pixels
Icon B moves right by about 11.25 pixels, and it still stays on the screen (its right edge is at about 361 < 800).
In simple words: Both icons fit on the screen but currently overlap. Sliding Icon B about 11 pixels to the right makes the two circles just touch.

Exam Tip: "Touching externally" always means distance between centres = sum of radii. Set up that equation first.