NCERT Solutions Class 9 Maths Chapter 1 Think and Reflect — Orienting Yourself: The Use of Coordinates
NCERT Class 9 Mathematics — Ganita Manjari Part I, Orienting Yourself: The Use of Coordinates. All 10 Think and Reflect prompts answered.
Also here. Every exercise in this chapter is solved, one page per exercise.
Ganita Manjari sets short Think and Reflect boxes into the body of each chapter. They are not exercises and they carry no marks, but they are where the chapter’s ideas are actually tested — usually by asking why something just worked, immediately after it worked. Chapter 1 has four such boxes, holding ten prompts between them, on printed pages $5$, $7$, $9$ and $11$.
Several of them are deliberately open: they ask you to look around your home, or to justify a claim rather than compute a number. Those are answered here with the reasoning set out in full, and with the measurement left where the book wants it — with you.
Key insight. Three of these four boxes are asking the same question in different clothing: what does a coordinate actually mean? A zero coordinate says the point is on an axis. Swapping the two coordinates moves the point unless they were equal. Changing the sign of one coordinate reflects the point across the other axis. Each of these is a sentence about geometry and a sentence about arithmetic at the same time, and being able to move between the two is what the chapter is for.
Prompts 1 and 2 come from the box on page 5, printed just after Exercise Set 1.1, where the door to Reiaan’s room was found to be $3.5$ ft wide.
Question 1
What are the standard widths for a room door? Look around your home and in school.
Solution. This is a measuring task, and the measuring is the point — but it helps to know roughly what to expect, so that a tape measure that says something very different prompts a second look.
In Indian homes the usual internal door is about $0.75$ m to $0.90$ m wide, which is $2.5$ ft to $3$ ft. Main entrance doors are wider, commonly $1.0$ m to $1.2$ m. Bathroom doors are often the narrowest in the house, sometimes only $0.6$ m to $0.75$ m — which is exactly the pattern the floor plan in Exercise Set 1.1 showed, with a $3.5$ ft room door and a $2.5$ ft bathroom door.
Measure the clear opening — the gap you actually walk through when the door is fully open — not the frame. The two differ by an inch or two, and it is the clear opening that decides whether something fits through.
Typical widths: internal room doors about $0.75$–$0.90$ m ($2.5$–$3$ ft), main doors about $1.0$–$1.2$ m, bathroom doors often $0.6$–$0.75$ m.
Measure the clear opening rather than the frame, and compare your own doors against these figures.
Question 2
Are the doors in your school suitable for people in wheelchairs?
Solution. Turn the question into a comparison of two numbers: how wide is the clear opening, and how much width does a wheelchair need?
A standard manually propelled wheelchair is about $0.65$ m to $0.70$ m across at the wheels, and accessibility guidance generally asks for roughly $0.90$ m of clear opening so that a person can pass without scraping their hands on the frame. So a door with a clear opening below about $0.80$ m is likely to be a problem, and one at $0.90$ m or more is comfortable.
Width is not the whole of it, though, and a careful answer says so. A door can be wide enough and still be unusable if there is a step at the threshold, if the corridor outside is too narrow to turn in, if the door is heavy or spring-loaded, or if the handle is set too high. The door in Exercise Set 1.1 passed the width test easily at $3.5$ ft; whether the route to it works is a separate question, and one the floor plan cannot answer.
Compare each door’s clear opening with about $0.90$ m. A wheelchair is roughly $0.65$–$0.70$ m wide, so anything under about $0.80$ m will be difficult.
Width alone is not enough: check also for a step at the threshold, the space to turn in the corridor outside, how heavy the door is, and the height of the handle.
Prompts 3 to 6 come from the box on page 7, after the discussion of the four quadrants and just before Exercise Set 1.2.
Question 3
What is the $x$-coordinate of a point on the $y$-axis?
Solution. The $x$-coordinate measures how far the point is from the $y$-axis. A point that lies on the $y$-axis is no distance from it at all.
$0$. Every point on the $y$-axis has the form $(0, y)$.
Question 4
Is there a similar generalisation for a point on the $x$-axis?
Solution. Yes, and by exactly the same reasoning with the roles swapped. The $y$-coordinate measures distance from the $x$-axis, so a point sitting on the $x$-axis has $y = 0$.
Putting the two together explains the origin: it is the one point lying on both axes, so both of its coordinates are zero — which is what Question 1 of the End-of-Chapter Exercises asks.
Yes. Every point on the $x$-axis has the form $(x, 0)$ — its $y$-coordinate is $0$.
The origin $(0, 0)$ is the single point satisfying both conditions at once.
Question 5
Does point $Q(y, x)$ ever coincide with point $P(x, y)$? Justify your answer.
Solution. The two points coincide when their first coordinates agree and their second coordinates agree — both conditions, not one. Writing that out:
$$\text{first coordinates: } y = x \qquad \text{second coordinates: } x = y$$
Both say the same thing. So $P$ and $Q$ are the same point precisely when $x = y$, and otherwise they are two different points.
They do coincide, then, but only for a special family of points: those with equal coordinates, such as $(0,0)$, $(2,2)$, $(-5,-5)$ and $\left(\tfrac{1}{2}, \tfrac{1}{2}\right)$. Plot a few and you will see them lining up along the diagonal that runs through the origin between Quadrants I and III.
Geometrically, swapping the coordinates reflects a point in that diagonal — and the only points a reflection leaves where they were are the ones on the mirror line itself.
Yes, but only when $x = y$.
$P(x, y)$ and $Q(y, x)$ are the same point exactly when the two coordinates are equal, so the points that stay put are those on the diagonal line through the origin: $(0,0)$, $(3,3)$, $(-4,-4)$ and so on. For every other point, swapping the coordinates moves it.
Question 6
If $x \neq y$, then $(x, y) \neq (y, x)$; and $(x, y) = (y, x)$ if and only if $x = y$. Is this claim true?
Solution. It is true, and it is Question 5’s finding stated carefully. The claim has two halves, and each is worth checking separately.
If $x \neq y$ then $(x, y) \neq (y, x)$. Two points are equal only when both coordinates match. If the first coordinates matched we would have $x = y$, which is ruled out. So the points differ. For instance $(2, 7)$ and $(7, 2)$ are quite different places: one is in the upper part of Quadrant I, the other well to the right.
$(x, y) = (y, x)$ if and only if $x = y$. “If and only if” needs both directions. If $x = y$ the two pairs are literally the same pair, so the points coincide. Conversely, if the points coincide then matching the first coordinates gives $x = y$.
This is why a coordinate pair is called an ordered pair. The order carries information: $(2, 7)$ and $(7, 2)$ are built from the same two numbers and name different points. It is the same point being made in Question 14 of the End-of-Chapter Exercises, where the street crossings $(4, 3)$ and $(3, 4)$ are two different corners of the city.
Yes, the claim is true. Two points are equal only when both coordinates agree, so $(x, y) = (y, x)$ forces $x = y$; and if $x = y$ the two pairs are identical.
This is precisely what makes a coordinate pair ordered — $(2, 7)$ and $(7, 2)$ are different points.
Prompts 7 and 8 come from the box on page 9, printed under Fig. 1.6, which shows the acute-angled triangle $ADM$ with $A(3, 4)$, $D(7, 1)$ and $M(9, 6)$ in the first quadrant.
Question 7
In moving from $A(3, 4)$ to $D(7, 1)$, what distance has been covered along the $x$-axis? What about the distance along the $y$-axis?
Solution. Split the journey into two moves along the grid: first across, then down.
Across, the $x$-coordinate goes from $3$ to $7$:
$$7 – 3 = 4 \text{ units, to the right}$$
Down, the $y$-coordinate goes from $4$ to $1$:
$$|1 – 4| = 3 \text{ units, downwards}$$
The book marks the corner where these two moves meet as $C(3, 1)$, directly below $A$ and directly left of $D$.
$4$ units along the $x$-axis (from $x = 3$ to $x = 7$) and $3$ units along the $y$-axis (from $y = 4$ down to $y = 1$).
Question 8
Can these distances help you find the distance $AD$?
Solution. They can, and this is the moment the chapter has been building towards.
The two moves are at right angles to each other — one horizontal, one vertical — so $A$, $C(3,1)$ and $D$ form a right-angled triangle with the right angle at $C$. The direct distance $AD$ is its hypotenuse, and the Baudhāyana–Pythagoras theorem gives it:
$$AD = \sqrt{AC^2 + CD^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ units}$$
Notice what has happened. The horizontal and vertical distances are easy — each is a single subtraction of coordinates. The slanting distance looks harder but is not, because it is always the hypotenuse of the right-angled triangle those two easy distances build. That observation, generalised, is the distance formula:
$$PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$
Yes. The horizontal move ($4$) and the vertical move ($3$) are the two legs of a right-angled triangle whose hypotenuse is $AD$, so
$$AD = \sqrt{4^2 + 3^2} = 5 \text{ units}$$
Generalising this gives the distance formula $PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$.
Prompts 9 and 10 come from the box on page 11, after Fig. 1.9, where triangle $ADM$ has been reflected in the $y$-axis to give $A'(-3, 4)$, $D'(-7, 1)$ and $M'(-9, 6)$.
Question 9
What has remained the same and what has changed with this reflection?
Solution. Compare the two triangles piece by piece rather than in general terms.
What has stayed the same. Every side length is unchanged:
$$AD = A’D’ = 5, \qquad DM = D’M’ = \sqrt{29}, \qquad MA = M’A’ = \sqrt{40}$$
Since all three sides match, the two triangles are congruent, so the angles are equal and the area is the same — $13$ square units in both cases. Each point has also kept its $y$-coordinate: $A$ and $A’$ are both at height $4$, $D$ and $D’$ both at height $1$. In other words, every point has stayed exactly as far from the $x$-axis as it was.
What has changed. Each $x$-coordinate has changed sign: $3 \to -3$, $7 \to -7$, $9 \to -9$. The triangle has moved from the first quadrant to the second, and each point is now on the opposite side of the $y$-axis at the same distance from it.
There is one more change, and it is the one usually missed. Going round the original triangle $A \to D \to M$ takes you anticlockwise; going round $A’ \to D’ \to M’$ takes you clockwise. The sense has reversed. This is why a reflected triangle cannot be slid across the page onto its image — you would have to pick it up and turn it over, which is exactly what a mirror does.
Unchanged: all three side lengths ($5$, $\sqrt{29}$, $\sqrt{40}$), and therefore all three angles and the area; and every point’s $y$-coordinate, so every point’s distance from the $x$-axis.
Changed: the sign of every $x$-coordinate, so the triangle moves from Quadrant I to Quadrant II; and its orientation — $A \to D \to M$ runs anticlockwise while $A’ \to D’ \to M’$ runs clockwise, so the image is a mirror image, not a slide.
Question 10
Would these observations be the same if $\triangle ADM$ is reflected in the $x$-axis (instead of the $y$-axis)?
Solution. Yes — the same observations hold with the two coordinates exchanging roles.
Reflecting in the $x$-axis keeps each $x$-coordinate and changes the sign of each $y$:
$$A(3, 4) \to (3, -4), \qquad D(7, 1) \to (7, -1), \qquad M(9, 6) \to (9, -6)$$
Check the side lengths. From $(3,-4)$ to $(7,-1)$ the moves are $4$ across and $3$ up, so that side is still $5$. The other two work out to $\sqrt{29}$ and $\sqrt{40}$ as before. The reason is the same in both cases: the distance formula squares each difference, and squaring destroys the sign, so changing the sign of a coordinate cannot change any distance.
What differs is only the description. Reflecting in the $y$-axis preserved $y$-coordinates and moved the triangle into Quadrant II; reflecting in the $x$-axis preserves $x$-coordinates and moves it into Quadrant IV. The orientation reverses in this case too.
Yes, with $x$ and $y$ swapping roles. Reflecting in the $x$-axis gives $(3, -4)$, $(7, -1)$ and $(9, -6)$: the $x$-coordinates are unchanged, every $y$-coordinate changes sign, the triangle moves into Quadrant IV, all three side lengths are preserved, and the orientation reverses again.
Lengths survive either reflection because the distance formula squares each difference, and a square is blind to the sign.
Common mistakes
- Question 3 and Question 4, swapping the two rules. “A point on the $y$-axis has $y = 0$” is the commonest slip in the chapter. Anchor it to the meaning instead of memorising it: the $x$-coordinate is the distance from the $y$-axis, so lying on that axis makes it zero.
- Question 5, answering a flat “no”. The temptation is to say $(x, y)$ and $(y, x)$ are always different, because the pairs are written differently. They coincide for every point with $x = y$ — an infinite family, lying along the diagonal through the origin.
- Question 6, proving only one direction. “If and only if” is two claims. Showing that $x = y$ makes the points equal is the easy half; the claim is not established until the converse is shown too.
- Question 7, giving the answer as $-3$. A distance covered is not a coordinate difference with its sign attached. The $y$-coordinate changes by $-3$; the distance travelled is $3$ units, downwards. Saying which direction is better than a minus sign.
- Question 8, adding the two distances. Treating $AD$ as $4 + 3 = 7$ measures the path along the grid — across, then down — not the straight line. The straight line is always shorter, and here it is $5$.
- Question 9, listing only the coordinates that changed. The question asks for both halves. The genuinely interesting answer is what survived: the lengths, and therefore the shape itself.
- Question 9, missing the reversal of orientation. A reflected triangle is congruent to the original but cannot be slid onto it. Checking the direction you travel round the vertices, $A \to D \to M$, is what reveals this, and it is the difference between a reflection and a rotation.
Practise next
- The End-of-Chapter Exercises, Question 3 — asks which two of four given points are mirror images in an axis, which is Questions 9 and 10 here read backwards.
- The End-of-Chapter Exercises, Questions 1 and 2 — both rest directly on Questions 3 and 4 above: what a coordinate of zero tells you about where a point lies.
- Exercise Set 1.2 — where the signs first have to be handled in a real setting, with the bathroom to the left of the $y$-axis and the dining room below the $x$-axis.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.