Home · Schooling · CBSE · Class 9 · Mathematics · Orienting Yourself: The Use of Coordinates · Exercise Set 1.2

NCERT Class 9 Mathematics Exercise Set 1.2 Solutions: Orienting Yourself: The Use of Coordinates

Step-by-step working for every question of Exercise Set 1.2 in NCERT's Class 9 Mathematics book, chapter “Orienting Yourself: The Use of Coordinates”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Ex 1.2 Q1

   Reiaan's rectangular study table has three feet at (8, 9), (11, 9) and (11, 7). (i) Where is the fourth foot? (ii) Is it a good spot (Fig. 1.5)? (iii) Width and length of the table; can the height be found?

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. (i) The feet (8, 9) and (11, 9) have the same y-coordinate, and (11, 9) and (11, 7) have the same x-coordinate, so the corner between them is (11, 9).
   2. (i) In a rectangle the fourth corner takes the x-coordinate of (8, 9) and the y-coordinate of (11, 7), so the fourth foot is at (8, 7).
   3. (ii) The table covers x from 8 to 11 and y from 7 to 9. The bed ends at x = 7 and the wardrobe is near the x-axis, so the table overlaps nothing.
   4. (ii) It is 1 ft from the top wall (y = 10) and 1 ft from the right wall (x = 12), away from both doors, so it is a reasonable spot; the gap of 1 ft to the bed is tight for a chair, so moving it a little is also a fair choice.
   5. (iii) The shorter side runs from y = 7 to y = 9: width = 9 − 7 = 2 ft.
   6. (iii) The longer side runs from x = 8 to x = 11: length = 11 − 8 = 3 ft.
   7. (iii) The coordinates only describe the floor (a 2-D plan), so the height of the table cannot be found from them.

   Answer: (i) (8, 7); (ii) yes, it fits clear of the bed, wardrobe and doors; (iii) width 2 ft, length 3 ft; the height cannot be found from a floor plan

   Report an error in this solution

2. Ex 1.2 Q2

   The bathroom door B₁B₂ (B₁ = (0, 1.5), B₂ = (0, 4)) is hinged at B₁ and opens into the bedroom. Will it hit the wardrobe W₁W₂W₃W₄ (W₁ = (3, 0), W₂ = (7, 0), W₃ = (7, 2), W₄ = (3, 2))? What if the door is made wider?

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. The door B₁B₂ is 4 − 1.5 = 2.5 ft wide, so as it swings about the hinge B₁ (0, 1.5) its free end moves on a quarter circle of radius 2.5 ft inside the bedroom.
   2. The wardrobe covers x from 3 to 7 and y from 0 to 2. Its point nearest to B₁ is on its left side x = 3, level with the hinge: (3, 1.5).
   3. Distance from B₁ (0, 1.5) to (3, 1.5) = 3 − 0 = 3 ft. Every other point of the wardrobe is farther away (for example W₄ (3, 2) is √(3² + 0.5²) = √9.25 ≈ 3.04 ft away).
   4. The door reaches only 2.5 ft from the hinge, and 2.5 ft < 3 ft, so it does not hit the wardrobe.
   5. If the door is made wider than 3 ft (and hinged at B₁ as now), its edge would reach the wardrobe, so a wider door should be hinged at B₂ instead, or open into the bathroom, or the wardrobe should be moved.

   Answer: The door (2.5 ft) does not hit the wardrobe, whose nearest point is 3 ft from the hinge; a door wider than 3 ft hinged at B₁ would hit it.

   Report an error in this solution

3. Ex 1.2 Q3

   Reiaan's bathroom (Fig. 1.5): (i) coordinates of its corners O, F, R, P; (ii) shape and corner coordinates of the showering area SHWR; (iii) mark a 3 ft × 2 ft washbasin space and a 2 ft × 3 ft toilet space and give their corners.

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. (i) Read from Fig. 1.5: O is the origin and F is on the y-axis at 9, so O = (0, 0) and F = (0, 9).
   2. (i) The bathroom's left wall is at x = −6: R = (−6, 9) at the top and P = (−6, 0) at the bottom. So the bathroom is a 6 ft × 9 ft rectangle.
   3. (ii) The showering area has corners S = (−6, 6), H = (−3, 6), W = (−2, 9) and R = (−6, 9).
   4. (ii) SH (y = 6) and RW (y = 9) are both horizontal, so SH ∥ RW, while HW is slanted. A quadrilateral with exactly one pair of parallel sides is a trapezium (here a right trapezium, since SR is vertical).
   5. (iii) One possible layout (other placements are also correct): a washbasin space 3 ft along the bottom wall and 2 ft deep at the bottom-left corner, with corners (−6, 0), (−3, 0), (−3, 2), (−6, 2).
   6. (iii) A toilet space 2 ft wide and 3 ft deep along the left wall just above it, with corners (−6, 2.5), (−4, 2.5), (−4, 5.5), (−6, 5.5); it stays below the showering area, which starts at y = 6.

   Answer: (i) O (0, 0), F (0, 9), R (−6, 9), P (−6, 0); (ii) a trapezium with S (−6, 6), H (−3, 6), W (−2, 9), R (−6, 9); (iii) one choice: washbasin (−6, 0), (−3, 0), (−3, 2), (−6, 2) and toilet (−6, 2.5), (−4, 2.5), (−4, 5.5), (−6, 5.5)

   Report an error in this solution

4. Ex 1.2 Q4

   (i) The dining room (18 ft long, 15 ft wide) is entered through Reiaan's room door; its length runs from P (−6, 0) to A (12, 0). Sketch it and give its corners. (ii) Place a 5 ft × 3 ft table exactly at its centre: coordinates of the table's feet.

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. (i) The dining room has length 18 ft and width 15 ft.
   2. The length runs from P(−6, 0) to A(12, 0), which is the top edge of the room.
   3. Since the room extends downwards from the x-axis, the bottom corners are directly below P and A by 15 ft.
   4. The bottom-right corner is (12, 0 − 15) = (12, −15).
   5. The bottom-left corner is (−6, 0 − 15) = (−6, −15).
   6. Thus, the four corners of the dining room are (−6, 0), (12, 0), (12, −15), and (−6, −15).
   7. (ii) The table is 5 ft long and 3 ft wide, placed exactly at the centre of the room.
   8. The centre of the room is the midpoint of the diagonal from (−6, 0) to (12, −15).
   9. Centre x-coordinate = (−6 + 12) ÷ 2 = 3.
   10. Centre y-coordinate = (0 + (−15)) ÷ 2 = −7.5.
   11. So, the centre of the room is (3, −7.5).
   12. The table is centred at (3, −7.5) with length 5 ft (along x) and width 3 ft (along y).
   13. Half-length of table = 5 ÷ 2 = 2.5 ft; half-width = 3 ÷ 2 = 1.5 ft.
   14. Left x-coordinate of table = 3 − 2.5 = 0.5; Right x-coordinate = 3 + 2.5 = 5.5.
   15. Top y-coordinate of table = −7.5 + 1.5 = −6; Bottom y-coordinate = −7.5 − 1.5 = −9.
   16. The four feet of the table are at (0.5, −6), (5.5, −6), (5.5, −9), and (0.5, −9).

   Answer: (i) Corners: (−6, 0), (12, 0), (12, −15), (−6, −15); (ii) Table feet: (0.5, −6), (5.5, −6), (5.5, −9), (0.5, −9); centre (3, -7.5)

   Report an error in this solution

← Orienting Yourself: The Use of Coordinates: notes, practice and all exercises