NCERT Solutions for Class 9 Maths Chapter 1 — Exercise Set 1.2#

Call the three feet $P_1(8, 9)$, $P_2(11, 9)$ and $P_3(11, 7)$.

• $P_1P_2$ is horizontal, because both points have $y = 9$.
• $P_2P_3$ is vertical, because both points have $x = 11$.

So $P_1$, $P_2$ and $P_3$ are three corners of a rectangle. The fourth corner has the same $x$-coordinate as $P_1$ and the same $y$-coordinate as $P_3$:

$$P_4 = (8, 7)$$

∴ The fourth foot is at (8, 7).

The table covers $8 \leq x \leq 11$ and $7 \leq y \leq 9$. In Fig. 1.5, the bed ends at about $x = 7$ and the right wall is at $x = 12$, so the table sits in the free space between the bed and the right wall, inside the room, without overlapping any furniture.

∴ Yes, this is a reasonable spot for the study table.

The sides of the table are parallel to the axes, so we subtract the coordinates that change.

Side from $(8, 9)$ to $(11, 9)$:

$$\text{Side} = 11 - 8 = 3 \text{ ft}$$

Side from $(11, 9)$ to $(11, 7)$:

$$\text{Side} = 9 - 7 = 2 \text{ ft}$$

∴ The length is 3 ft (the longer side) and the width is 2 ft.

The height cannot be found from the figure. Fig. 1.5 is a top view of the room and shows only two coordinates, $x$ and $y$. Height would need a third direction, which the plan does not show.

The bathroom door runs from $B_1(0, 1.5)$ to $B_2(0, 4)$:

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

When the door swings into the bedroom about the hinge $B_1$, its free end moves on a quarter-circle of radius 2.5 ft around $B_1$. The farthest it can reach from the y-axis is 2.5 ft, at the point $(2.5, 1.5)$.

The wardrobe starts at $W_1(3, 0)$, so its nearest side is the line $x = 3$, for $0 \leq y \leq 2$:

$$\Rightarrow 2.5 < 3$$

∴ The door reaches only 2.5 ft from the wall, which is less than 3 ft, so the door will not hit the wardrobe.

If the door is made wider: the point $(3, 1.5)$ on the wardrobe is exactly 3 ft from the hinge $B_1(0, 1.5)$. So a door 3 ft wide or wider would hit the wardrobe. If the door must be wider, we could:

• move the wardrobe further to the right, away from the y-axis;
• put the hinge at $B_2$ instead, so that the door swings upwards, away from the wardrobe; or
• make the door open into the bathroom.

Reading from Fig. 1.5:

Corner	Coordinates	Position
$O$	$(0, 0)$	bottom-right (the origin)
$F$	$(0, 9)$	top-right (on the y-axis)
$R$	$(-6, 9)$	top-left
$P$	$(-6, 0)$	bottom-left

The bathroom is a rectangle 6 ft wide (from $x = -6$ to $x = 0$) and 9 ft long (from $y = 0$ to $y = 9$).

Reading from Fig. 1.5:

Corner	Coordinates
$S$	$(-6, 6)$
$H$	$(-3, 6)$
$W$	$(-2, 9)$
$R$	$(-6, 9)$

We look at each side:

• $SH$ joins $(-6, 6)$ and $(-3, 6)$. Both have $y = 6$, so $SH$ is horizontal, and $SH = -3 - (-6) = 3$ ft.
• $RW$ joins $(-6, 9)$ and $(-2, 9)$. Both have $y = 9$, so $RW$ is horizontal, and $RW = -2 - (-6) = 4$ ft.
• $SR$ joins $(-6, 6)$ and $(-6, 9)$. Both have $x = -6$, so $SR$ is vertical, and $SR = 9 - 6 = 3$ ft.
• $HW$ joins $(-3, 6)$ and $(-2, 9)$. The $x$-coordinate changes from $-3$ to $-2$, so $HW$ is slanted.

$SH$ and $RW$ are both horizontal, so they are parallel, but they have different lengths (3 ft and 4 ft). $SR$ is perpendicular to both.

∴ The showering area $SHWR$ is a trapezium (a right trapezium, with right angles at $S$ and $R$).

Answers can vary. The spaces must lie inside the bathroom ($-6 \leq x \leq 0$, $0 \leq y \leq 9$) and below the showering area (below $y = 6$). Following Fig. 1.5, we put both against the left wall $x = -6$.

Washbasin, 3 ft × 2 ft (3 ft along the wall, 2 ft into the room), in the lower part:

Corner	Coordinates
bottom-left	$(-6, 0)$
bottom-right	$(-4, 0)$
top-right	$(-4, 3)$
top-left	$(-6, 3)$

Toilet, 2 ft × 3 ft (2 ft along the wall, 3 ft into the room), above the washbasin:

Corner	Coordinates
bottom-left	$(-6, 3.5)$
bottom-right	$(-3, 3.5)$
top-right	$(-3, 5.5)$
top-left	$(-6, 5.5)$

Both spaces lie inside the bathroom, below the showering area, and do not overlap each other.

From Fig. 1.5, $P = (-6, 0)$ and $A = (12, 0)$. Both lie on the x-axis, so:

$$PA = 12 - (-6) = 18 \text{ ft}$$

This matches the length of the dining room. The door $D_1R_1$ of Reiaan’s room is on the x-axis, so the dining room lies on the other side of the x-axis, below it. Its width is 15 ft, so it goes down to $y = -15$.

∴ The corners of the dining room are $(-6, 0)$, $(12, 0)$, $(12, -15)$ and $(-6, -15)$.

These fit on the graph sheet, which marks the y-axis down to $(0, -15)$.

The centre of the dining room is halfway across and halfway down:

$$x\text{-coordinate of centre} = \dfrac{-6 + 12}{2} = 3$$

$$y\text{-coordinate of centre} = \dfrac{0 + (-15)}{2} = -7.5$$

So the centre is $(3, -7.5)$. We place the 5 ft side along the x-direction and the 3 ft side along the y-direction. The feet are 2.5 ft to the left and right of the centre, and 1.5 ft above and below it:

$$x = 3 - 2.5 = 0.5 \quad \text{or} \quad x = 3 + 2.5 = 5.5$$

$$y = -7.5 - 1.5 = -9 \quad \text{or} \quad y = -7.5 + 1.5 = -6$$

∴ The feet of the table are at $(0.5, -9)$, $(5.5, -9)$, $(5.5, -6)$ and $(0.5, -6)$.