NCERT Solutions for Class 9th Maths Chapter 1 Exercise Set 1.2 — The 2-d Cartesian Coordinate System

Book page 8 Updated on2026-09-08

Q1.

Place Reiaan’s rectangular study table with three of its feet at the points (8, 9), (11, 9) and (11, 7). (i) Where will the fourth foot of the table be? (ii) Is this a good spot for the table? (iii) What is the width of the table? The length? Can you make out the height of the table?

Answer

(i) The fourth foot is at (8, 7).

(8, 9) and (11, 9) share the same y → that side is horizontal
(11, 9) and (11, 7) share the same x → that side is vertical
The fourth vertex must share x = 8 with (8, 9) and y = 7 with (11, 7)
Fourth foot = (8, 7)

Why it happens: In a rectangle the opposite sides are parallel and equal. The shift from (11, 9) to (8, 9) is 3 units left, so the same shift applied to (11, 7) must land on the fourth foot: (11 − 3, 7) = (8, 7). You never need to measure — the shift does the work.

(ii) Yes, it is a workable spot, and it can be improved.

The table covers 8 ≤ x ≤ 11 and 7 ≤ y ≤ 9. Check it against everything else in the room:

Bed: x from 0.5 to 6.5. Since 8 > 6.5, no clash.
Wardrobe: x from 3 to 7, y from 0 to 2. No clash.
Room door D₁R₁: x from 8 to 11.5 on the bottom wall. The table is at y ≥ 7, far from the swing of the door.

So nothing overlaps. But it leaves a wasted 1 ft gap on the right (x = 11 to the right wall at x = 12) and 1 ft above (y = 9 to the top wall at y = 10). Pushing the table into the corner, with feet at (9, 8), (12, 8), (12, 10) and (9, 10), would clear more floor and — important for Reiaan — put two of its sides against walls he can trace with his hand.

(iii) Width 2 ft, length 3 ft; the height cannot be found.

Length = |11 − 8| = 3 ft (along the x-direction)
Width = |9 − 7| = 2 ft (along the y-direction)
Height = cannot be determined

Why the height is missing: Fig. 1.5 is a plan of the floor. Its two axes fix the two directions that lie in the floor; height is perpendicular to both and no pair (x, y) can record it. This is the same reason the windows could not be marked on Fig. 1.1.

Q2.

If the bathroom door has a hinge at B₁ and opens into the bedroom, will it hit the wardrobe? Are there any changes you would suggest if the door is made wider?

Answer

No, it will not hit the wardrobe — but the clearance is only 0.5 ft.

Hinge B₁ = (0, 1.5); door leaf B₁B₂ = |4 − 1.5| = 2.5 ft
Swinging on the hinge, the free edge traces a quarter circle of radius 2.5 about B₁
Wardrobe occupies 3 ≤ x ≤ 7, 0 ≤ y ≤ 2
Nearest point of the wardrobe to B₁ is (3, 1.5), because y = 1.5 already lies between 0 and 2
Distance = |3 − 0| = 3 ft
3 ft > 2.5 ft → the door misses the wardrobe by 0.5 ft

Why it happens: Every point of the swinging door stays within 2.5 ft of the hinge, because the hinge holds one end fixed. So the whole question reduces to one comparison: is the nearest bit of the wardrobe more than 2.5 ft from B₁? Turning “does it hit?” into “compare two distances” is exactly what coordinates are for.

If the door is made wider. A door of width w hinged at B₁ will just graze the wardrobe when w = 3 ft, and will strike it for any w > 3 ft. Since 2.5 ft (30 in) is below the 32-inch clear width that wheelchair users need, widening the door is worth doing — so make one of these changes:

Move the wardrobe 0.5–1 ft to the right, from W₁ (3, 0)–W₂ (7, 0) to (3.5, 0)–(7.5, 0). There is room: the right wall is at x = 12.
Hinge the door at B₂ (0, 4) instead, so it swings upward, away from the wardrobe. The nearest wardrobe corner is then (3, 2), at distance √(3² + 2²) = √13 ≈ 3.6 ft — so a door up to about 3.5 ft would clear.
Make it open into the bathroom, or fit a sliding door, which sweeps no floor at all.

Tip: The door swing is dead floor space — you cannot put anything there. On a plan, sketch each door’s quarter circle before you place the furniture, not after.

Q3.

Look at Reiaan’s bathroom. (i) What are the coordinates of the four corners O, F, R, and P of the bathroom? (ii) What is the shape of the showering area SHWR in Reiaan’s bathroom? Write the coordinates of the four corners. (iii) Mark off a 3 ft × 2 ft space for the washbasin and a 2 ft × 3 ft space for the toilet. Write the coordinates of the corners of these spaces.

Answer

(i) Reading Fig. 1.5, the bathroom lies to the left of the y-axis:

O = (0, 0), F = (0, 9), R = (−6, 9), P = (−6, 0)
Check: width = |0 − (−6)| = 6 ft, length = |9 − 0| = 9 ft → a 6 ft × 9 ft bathroom, as Fig. 1.1 states

Why the x-coordinates are negative: The bathroom is on the far side of the wall that was chosen as the y-axis. Distances to the left of that axis are counted as negative, so the bathroom sits in Quadrant II. This is the whole payoff of allowing negative numbers: one origin and one pair of axes can serve both rooms.

(ii) The showering area is a trapezium (a right trapezium), not a rectangle.

The showering area SHWR inside the 6 ft × 9 ft bathroom. SH and RW are parallel; RS is perpendicular to both.

S = (−6, 6), H = (−3, 6), W = (−2, 9), R = (−6, 9)
SH: from (−6, 6) to (−3, 6) — both have y = 6, so SH is horizontal, length 3 ft
RW: from (−6, 9) to (−2, 9) — both have y = 9, so RW is horizontal, length 4 ft
SH ∥ RW but 3 ≠ 4 → exactly one pair of parallel sides → trapezium
RS: from (−6, 9) to (−6, 6) — both have x = −6, so RS is vertical, length 3 ft, and is perpendicular to SH and RW
HW = √((−2 − (−3))² + (9 − 6)²) = √(1 + 9) = √10 ≈ 3.16 ft — the slanting glass edge
Area = ½ × (3 + 4) × 3 = 10.5 sq ft

Why we can tell the shape without measuring: Two segments are parallel to each other when the same coordinate is constant along both. Here y is constant along SH and along RW, so both are horizontal; x is constant along RS, so it is vertical and therefore perpendicular to them. HW has neither coordinate constant, so it is slanted — and its length needs the Baudhāyana–Pythagoras Theorem.

(iii) Answers may differ. One workable arrangement, keeping both fittings clear of the shower (which occupies y ≥ 6 on the left) and leaving a clear path from the doorway on the y-axis:

Fitting	Size	Corners
Washbasin	3 ft × 2 ft	(−6, 0), (−3, 0), (−3, 2), (−6, 2)
Toilet	2 ft × 3 ft	(−6, 2.5), (−4, 2.5), (−4, 5.5), (−6, 5.5)

Washbasin: 3 units wide along x (−6 to −3), 2 units along y (0 to 2) → 3 ft × 2 ft ✓
Toilet: 2 units along x (−6 to −4), 3 units along y (2.5 to 5.5) → 2 ft × 3 ft ✓
Both lie inside −6 ≤ x ≤ 0 and 0 ≤ y ≤ 9, and both stay below y = 6 → clear of the shower

Check it yourself: Whatever corners you choose, verify them the same way — subtract the x-coordinates to get one side, subtract the y-coordinates to get the other, and check that all four corners satisfy −6 ≤ x ≤ 0 and 0 ≤ y ≤ 9. Also leave the strip near x = 0, y = 1.5 to 4 free: that is where the doorway is.

Q4.

Other rooms in the house: (i) Reiaan’s room door leads from the dining room which has the length 18 ft and width 15 ft. The length of the dining room extends from point P to point A. Sketch the dining room and mark the coordinates of its corners. (ii) Place a rectangular 5 ft × 3 ft dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.

Answer

(i) P = (−6, 0) and A = (12, 0) are already on the plan, and

PA = |12 − (−6)| = 18 ft — exactly the stated length ✓

So PA is the wall shared with the bedroom and bathroom, and the dining room must lie on the other side of it — below the x-axis. Going 15 ft down from PA gives the corners:

P (−6, 0), A (12, 0), (12, −15), (−6, −15)

The dining room (blue) lies below the x-axis, sharing the wall PA with the bedroom and bathroom. The dining table (amber) is centred at (3, −7.5).

Why the y-coordinates are negative: The x-axis is the wall that separates the dining room from Reiaan’s side of the house. Distances measured downwards from that wall are negative by the sign convention, so the dining room occupies parts of Quadrants III and IV. That is also why the exercise told you to mark the y-axis all the way down to (0, −15) — the graph has to be large enough to hold the room.

(ii) The centre of a rectangle is the midpoint of a diagonal.

Centre = midpoint of P (−6, 0) and (12, −15)
= ((−6 + 12)/2, (0 + (−15))/2)
= (3, −7.5)

Table 5 ft along x and 3 ft along y → go 2.5 each side in x and 1.5 each side in y:
x: 3 − 2.5 = 0.5 and 3 + 2.5 = 5.5
y: −7.5 − 1.5 = −9 and −7.5 + 1.5 = −6
Feet = (0.5, −6), (5.5, −6), (5.5, −9), (0.5, −9)

Check: the midpoint of the table’s own diagonal is ((0.5 + 5.5)/2, (−6 + (−9))/2) = (3, −7.5) ✓ — the table really is centred.

Try This: Turn the table through a right angle, 3 ft along x and 5 ft along y. The feet become (1.5, −5), (4.5, −5), (4.5, −10), (1.5, −10) — still centred at (3, −7.5). Both answers are correct; the question fixes the centre, not the orientation.

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