
Exercise Set 1.2 Class 9 · Ganita Manjari
The given three points form three corners of a rectangle:
To complete the rectangle, the fourth point must have:
- the same x‑coordinate as A → 8
- the same y‑coordinate as C → 7
Yes, this is a good spot because:
- The table is placed neatly inside the room.
- It does not block doors or pathways.
- It is positioned near the wall, which is practical for study.
Width:
Distance between (8, 9) and (11, 9) = 11 − 8 = 3 units
Length:
Distance between (11, 9) and (11, 7) = 9 − 7 = 2 units
Height:
The height of the table cannot be determined from the given diagram because the figure represents only a top view (2D), not vertical dimensions.
From Fig. 1.5:
So, the bathroom door has width:
= 4 – 1.5 = 2.5 units
If the door is hinged at B₁ and opens into the bedroom, it will sweep an arc of radius 2.5 units from B₁.
Now the wardrobe begins at:
The nearest point of the wardrobe from B₁(0, 1.5) is around x = 3, which is farther than the door width 2.5 units.
Suggestion if the door is made wider:
- If the door becomes much wider, it may come close to or hit the wardrobe.
- In that case, the door could be made to open inward into the bathroom, or
- the wardrobe could be shifted slightly to the right, or
- the door width could be kept limited for comfortable movement.
From Fig. 1.5, the coordinates of the four corners O, F, R and P of the bathroom are:
From the figure:
Since one pair of opposite sides is parallel, SHWR is a trapezium.
Washbasin space (3 ft × 2 ft):
Take the rectangle near the bottom-left corner of the bathroom.
Toilet space (2 ft × 3 ft):
Take a rectangle above the washbasin.
From Fig. 1.5:
So, the length of PA = 12 – (−6) = 18 ft, which matches the given length.
If the dining room is 15 ft wide and lies below PA, then its upper side is PA and it extends downward 15 units.
Hence the coordinates of four corners are:
The dining room extends:
- From x = −6 to x = 12
- From y = 0 to y = −15
Centre of the dining room:
x‑coordinate of centre = (−6 + 12) / 2 = 3
y‑coordinate of centre = (0 + (−15)) / 2 = −7.5
Now place a 5 ft × 3 ft table at the centre.
Taking length = 5 units along the x‑axis and width = 3 units along the y‑axis:
Half‑length = 2.5 | Half‑width = 1.5
So the coordinates of corners (feet) of the table are: