NCERT Solutions Class 9 Maths Exercise 1.2 — Orienting Yourself: The Use of Coordinates

NCERT Class 9 Mathematics — Ganita Manjari Part I, Orienting Yourself: The Use of Coordinates, Exercise Set 1.2. All 4 questions solved.

Exercise Set 1.1 stayed inside Reiaan’s bedroom, where every coordinate was positive. Fig. 1.5 now extends the same plan in both directions — the bathroom lies to the left of the $y$-axis and the dining room below the $x$-axis — so this exercise is where negative coordinates start doing real work.

The plan gives these fixed points, and it is worth writing them down before starting:

• The bedroom $OABC$: $O(0, 0)$, $A(12, 0)$, $B(12, 10)$, $C(0, 10)$.
• The wardrobe: $W_1(3, 0)$, $W_2(7, 0)$, $W_3(7, 2)$, $W_4(3, 2)$.
• The bathroom door, on the $y$-axis: $B_1(0, 1.5)$ and $B_2(0, 4)$.
• The room door, on the $x$-axis: $D_1(8, 0)$ and $R_1(11.5, 0)$.

One unit is one foot throughout.

Key insight. A rectangle drawn with its sides along the axes is completely described by two numbers on each axis. Its four corners are every combination of those numbers, which is why three feet of a rectangular table determine the fourth, and why “$5$ ft $\times$ $3$ ft, centred at a point” fixes all four corners exactly. Whenever a question here looks as though it needs measuring, look instead for the two $x$-values and the two $y$-values that bound the shape.

Question 1

Place Reiaan’s rectangular study table with three of its feet at the points $(8, 9)$, $(11, 9)$ and $(11, 7)$.

Solution. Plot the three given feet first. Two of them, $(8, 9)$ and $(11, 9)$, share the $y$-value $9$, so that side is horizontal. Two of them, $(11, 9)$ and $(11, 7)$, share the $x$-value $11$, so that side is vertical. The table is therefore bounded by $x = 8$ and $x = 11$ on one pair of sides, and by $y = 7$ and $y = 9$ on the other.

(i) Where will the fourth foot of the table be?

The corners of an axis-parallel rectangle are the four pairings of its two $x$-values with its two $y$-values: $(8, 9)$, $(11, 9)$, $(11, 7)$ and the one still missing, $(8, 7)$.

(ii) Is this a good spot for the table?

Check what else occupies that patch of floor. The table covers the region $8 \le x \le 11$, $7 \le y \le 9$ — the top-right part of the room.

• The bed lies between $y = 5$ and $y = 8$, reaching from the left wall to about $x = 7$. The table starts at $x = 8$, so it is clear of the bed.
• The wardrobe lies below $y = 2$. No conflict.
• The room door runs along the bottom wall from $x = 8$ to $x = 11.5$, and it opens at floor level, $y = 0$. The table is seven feet further up the room, so it neither blocks the doorway nor stands in the path from the door to the bed.
• The table leaves a foot of clearance to the right wall ($x = 12$) and a foot to the top wall ($y = 10$) — enough to walk round two sides of it.

So geometrically it is a sensible spot: tucked into a corner, out of the traffic between the two doors, and not overlapping any furniture. What the plan cannot tell you is where the light and the power point are, and those matter for a study table — so the honest answer names the geometry and admits what the drawing leaves out.

(iii) What is the width of the table? The length? Can you make out the height of the table?

The two sides are measured as differences along each axis:

$$\text{along } x: \; 11 – 8 = 3 \text{ ft} \qquad \text{along } y: \; 9 – 7 = 2 \text{ ft}$$

So the table is $3$ ft long and $2$ ft wide.

The height is a different matter. This figure is a plan — a view from directly above — and it carries exactly two axes. Height is measured in the direction pointing out of the page, and there is no axis for it, so no amount of reading the figure can produce it. A two-dimensional coordinate system can locate a point on the floor; it cannot say anything about how far above the floor something rises.

Question 2

If the bathroom door has a hinge at $B_1$ and opens into the bedroom, will it hit the wardrobe? Are there any changes you would suggest if the door is made wider?

Solution. The door is the segment $B_1B_2$, of length $4 – 1.5 = 2.5$ ft, hinged at $B_1(0, 1.5)$. As it swings into the bedroom, its free end traces an arc of a circle centred on the hinge with radius $2.5$ ft. Nothing further than $2.5$ ft from $B_1$ can ever be touched by it.

So the question becomes: how close does the wardrobe come to $B_1$?

The wardrobe occupies $3 \le x \le 7$, $0 \le y \le 2$. Its left face is the segment $x = 3$ with $y$ running from $0$ to $2$. The hinge is at height $y = 1.5$, and $1.5$ lies inside that range, so the nearest point of the wardrobe is the point at the same height, straight across the floor:

$$\text{nearest point} = (3,\ 1.5), \qquad \text{distance} = \sqrt{(3-0)^2 + (1.5-1.5)^2} = 3 \text{ ft}$$

The door reaches $2.5$ ft; the wardrobe begins at $3$ ft.

$$3 – 2.5 = 0.5 \text{ ft}$$

Question 3

Look at Reiaan’s bathroom.

(i) What are the coordinates of the four corners $O$, $F$, $R$ and $P$ of the bathroom?

The bathroom is the region to the left of the $y$-axis. Its right wall is the $y$-axis itself, running from $O$ at floor level up to $F$; its left wall is the line $x = -6$, running from $P$ up to $R$. Reading the marked points from Fig. 1.5:

(ii) What is the shape of the showering area SHWR in Reiaan’s bathroom? Write the coordinates of the four corners.

Read the four marked corners from the figure: $S$ on the left wall at height $6$, $H$ on the $-3$ tick at the same height, $W$ on the $-2$ tick at height $9$, and $R$ the bathroom’s own top-left corner.

$$S(-6,\ 6) \qquad H(-3,\ 6) \qquad W(-2,\ 9) \qquad R(-6,\ 9)$$

Now test the sides. $S$ and $H$ share the $y$-value $6$, so $SH$ is horizontal; $R$ and $W$ share the $y$-value $9$, so $RW$ is horizontal too. Two horizontal lines are parallel, so $SH \parallel RW$. Their lengths differ, though:

$$SH = -3 – (-6) = 3 \qquad RW = -2 – (-6) = 4$$

The other two sides are not parallel — $SR$ is vertical, while $HW$ slopes from $(-3, 6)$ up to $(-2, 9)$. A quadrilateral with exactly one pair of parallel sides is a trapezium.

Its height is the gap between the two parallel sides, $9 – 6 = 3$ ft, so as a bonus:

$$\text{area} = \tfrac{1}{2}(3 + 4)(3) = 10.5 \text{ sq ft}$$

(iii) Mark off a $3$ ft $\times$ $2$ ft space for the washbasin and a $2$ ft $\times$ $3$ ft space for the toilet. Write the coordinates of the corners of these spaces.

Here the answer is not unique — any placement works provided each rectangle lies inside the bathroom ($-6 \le x \le 0$, $0 \le y \le 9$) and stays clear of the showering area and of the doorway at $B_1B_2$. Fig. 1.5 draws the washbasin in the bottom-left corner and the toilet above it against the left wall, so a placement matching the picture is:

Washbasin, $3$ ft along the wall and $2$ ft deep:

$$(-6,\ 0), \quad (-3,\ 0), \quad (-3,\ 2), \quad (-6,\ 2)$$

Toilet, $2$ ft along the wall and $3$ ft deep:

$$(-6,\ 3), \quad (-4,\ 3), \quad (-4,\ 6), \quad (-6,\ 6)$$

Check both: every $x$ lies between $-6$ and $0$, every $y$ between $0$ and $9$, the two rectangles do not overlap each other (one ends at $y = 2$, the other begins at $y = 3$), and the toilet stops exactly at $y = 6$, where the showering area begins.

Question 4

Other rooms in the house.

(i) Reiaan’s room door leads from the dining room, which has 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.

The door to Reiaan’s room is on the bottom wall, so the dining room must lie below the $x$-axis. Its length is the segment $PA$, and the two points are already known:

$$P(-6,\ 0) \qquad A(12,\ 0) \qquad PA = 12 – (-6) = 18 \text{ ft}$$

which agrees with the stated length. The width, $15$ ft, is measured downward from that edge, so the far wall is the line $y = -15$.

$$\text{Corners:} \quad P(-6,\ 0), \quad A(12,\ 0), \quad (12,\ -15), \quad (-6,\ -15)$$

This is also why the exercise told you to mark the $y$-axis down to $(0, -15)$: the dining room is exactly what fills that space. Its corners sit in the third quadrant ($(-6, -15)$) and the fourth ($(12, -15)$).

(ii) Place a rectangular $5$ ft $\times$ $3$ ft dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.

“Precisely in the centre” fixes the answer. The centre of a rectangle is the midpoint of either diagonal — the average of the two $x$-values and the average of the two $y$-values:

$$\text{centre} = \left( \frac{-6 + 12}{2},\ \frac{0 + (-15)}{2} \right) = \left( 3,\ -7.5 \right)$$

Put the $5$ ft side along the $x$-direction and the $3$ ft side along the $y$-direction. The table then reaches $2.5$ ft either side of the centre horizontally and $1.5$ ft either side vertically:

$$x: \; 3 \pm 2.5 \;\Rightarrow\; 0.5 \text{ and } 5.5 \qquad y: \; -7.5 \pm 1.5 \;\Rightarrow\; -9 \text{ and } -6$$

Common mistakes

• Question 1(i), joining the three points in the order given. Reading $(8,9)$, $(11,9)$, $(11,7)$ as three consecutive corners and then “continuing the pattern” to $(8, 5)$ or $(14, 7)$. Pair the repeated coordinates instead: the rectangle uses the $x$-values $\{8, 11\}$ and the $y$-values $\{7, 9\}$, and the missing corner is the pairing not yet used.
• Question 1(iii), inventing a height. A number like “$2.5$ ft” cannot come from this figure at all. The plan has two axes; height needs a third. Saying so is the answer, not a failure to answer.
• Question 2, comparing the door width with the distance to the wardrobe’s corner $W_1(3, 0)$. That corner is $\sqrt{3^2 + 1.5^2} \approx 3.35$ ft from the hinge, which makes the clearance look like $0.85$ ft. The door sweeps a circle, so what matters is the nearest point of the wardrobe, and because the hinge’s height $y = 1.5$ lies within the wardrobe’s range $0 \le y \le 2$, that nearest point is $(3, 1.5)$ at exactly $3$ ft.
• Question 2, treating the door as a straight line that slides. The door is hinged. Its reach is a radius, not a sideways shift, so every obstruction has to be measured from $B_1$ and from nowhere else.
• Question 3(i), writing $R$ as $(-6, 10)$. The bedroom’s top wall is at $y = 10$ but the bathroom’s is at $y = 9$, level with $F$. The two rooms do not have the same depth, and the figure marks $F$ on the $+9$ tick.
• Question 3(ii), calling $SHWR$ a rectangle or a parallelogram. Only one pair of sides is parallel: $SH$ and $RW$ are horizontal, but $SR$ is vertical while $HW$ slants. Two sides being parallel and unequal — $3$ against $4$ — rules a parallelogram out at once.
• Question 4(i), placing the dining room above the $x$-axis or to the left. The clue is in the setup: the dining room leads to Reiaan’s door, which is on the bottom wall, and the exercise asks for the $y$-axis to be marked down to $-15$. Both point downward.
• Question 4(ii), centring the table on the origin. The centre of the dining room is $(3, -7.5)$, not $(0, 0)$. The origin is a corner of Reiaan’s bedroom, which is a different room altogether.

Practise next

• The End-of-Chapter Exercises, Questions 1 to 5 — the same reading of coordinates, now without a floor plan to lean on. Question 3 in particular asks which sides of a quadrilateral are parallel or perpendicular, exactly as Question 3(ii) did here.
• Exercise Set 1.1 — worth a second look if the signs in this exercise felt awkward. Everything there sits in the first quadrant, which makes it a good place to confirm that the rule for a distance along an axis is the same whichever quadrant you are in.