NCERT Solutions for Class 9 Maths Chapter 1 — Exercise Set 1.1#

In this exercise you read positions of objects in Reiaan’s room from the Cartesian plane. A point $(x, y)$ lies $x$ units from the y-axis and $y$ units from the x-axis; the origin $O = (0, 0)$ is the bottom-left corner of the room and $1$ unit $= 1$ ft.

Fig. 1.3 shows Reiaan’s room with points OABC marking its corners. The x- and y-axes are marked in the figure. Point O is the origin. Referring to Fig. 1.3, answer the following questions: #

From the figure the relevant readings are: $O = (0, 0)$, $A = (12, 0)$, $B = (12, 10)$, $C = (0, 10)$, room door end-points $D_1 = (8, 0)$ and $R_1 = (11.5, 0)$, bathroom door end-points $B_1 = (0, 1.5)$ and $B_2 = (0, 4)$.

(i) If D₁R₁ represents the door to Reiaan’s room, how far is the door from the left wall (the y-axis) of the room? How far is the door from the x-axis?

The door runs from $D_1 = (8, 0)$ to $R_1 = (11.5, 0)$.

  • The point of the door nearest to the left wall (y-axis) is $D_1$, whose $x$-coordinate is $8$.
    So the door is $8$ ft from the y-axis.
  • Every point of the door has $y$-coordinate $0$ — the door lies on the x-axis, i.e. $0$ ft from the x-axis.
(ii) What are the coordinates of D₁?

Reading from Fig. 1.3, $D_1$ is on the x-axis at $x = 8$.

$$D_1 = (8,\ 0)$$
(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 he/she be able to do so easily?

Width of the door $= $ distance $D_1R_1$. Both points lie on the x-axis, so

$$\text{Width} = 11.5 - 8 = 3.5 \text{ ft}.$$

Yes, $3.5$ ft is a comfortable width for a room door (standard doors are $2.5$–$3$ ft).

A standard wheelchair is about $2$–$2.5$ ft wide, so a person in a wheelchair can enter the room easily through a $3.5$ ft door.

(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?

Both $B_1$ and $B_2$ lie on the y-axis, so

$$\text{Width of bathroom door} = 4 - 1.5 = 2.5 \text{ ft}.$$

Width of the room door $= 3.5$ ft (from part (iii)).

$$\Rightarrow 2.5 < 3.5$$

∴ Width of bathroom door < width of room door, so the bathroom door is narrower than the room door, by 1 ft.

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