NCERT Solutions Class 9 Maths Chapter 1 End-of-Chapter Exercises — Orienting Yourself: The Use of Coordinates

NCERT Class 9 Mathematics — Ganita Manjari Part I, Orienting Yourself: The Use of Coordinates, End-of-Chapter Exercises. All 16 questions solved.

These sixteen questions gather up everything in the chapter and push it a little further than the two exercise sets did. Four ideas do almost all the work:

• Where a point lies. A point $(x, y)$ is in Quadrant I if both coordinates are positive, II if $x < 0$ and $y > 0$, III if both are negative, IV if $x > 0$ and $y < 0$. A point with a zero coordinate lies on an axis and in no quadrant at all.
• Distance. For any two points, $PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$, which reduces to a single subtraction when the segment is parallel to an axis.
• Midpoint. $M$ is the midpoint of $ST$ exactly when each of its coordinates is the average of the corresponding coordinates of $S$ and $T$: $M = \left( \dfrac{x_S + x_T}{2}, \dfrac{y_S + y_T}{2} \right)$.
• Reflection. Reflecting in the $x$-axis changes the sign of $y$; reflecting in the $y$-axis changes the sign of $x$.

Questions $6$ to $13$ carry an asterisk in the book, which marks them as enrichment — they are not assessed. They are also the most interesting questions here, and they are solved in full below.

Key insight. Almost every question in this set is answered by turning a geometric word into an arithmetic one. “Parallel to an axis” means one coordinate is shared. “Mirror image in the $x$-axis” means same $x$, opposite $y$. “Midpoint” means average. “On the same straight line” means the steps between the points are in proportion. Once the translation is made, no drawing is needed — which is the whole point of coordinates.

Question 1

What are the $x$-coordinate and $y$-coordinate of the point of intersection of the two axes?

Solution. The two axes meet at one point only, the origin. A point on the $y$-axis has $x = 0$, and a point on the $x$-axis has $y = 0$. The point of intersection lies on both axes at once, so it must satisfy both conditions together.

Question 2

Point $W$ has $x$-coordinate equal to $-5$. Can you predict the coordinates of point $H$ which is on the line through $W$ parallel to the $y$-axis? Which quadrants can $H$ lie in?

Solution. A line parallel to the $y$-axis is vertical, and every point on a vertical line has the same $x$-coordinate. The line through $W$ parallel to the $y$-axis is therefore the line $x = -5$, and $H$ lies somewhere on it.

That fixes one coordinate and leaves the other completely free: $H = (-5, y)$ for some $y$. So the coordinates cannot be predicted exactly — only the first of them can.

Now place it. Since $x = -5$ is negative:

• if $y > 0$, the point is left of the $y$-axis and above the $x$-axis — Quadrant II;
• if $y < 0$, it is left and below — Quadrant III;
• if $y = 0$, it is the point $(-5, 0)$, sitting on the negative $x$-axis, which belongs to no quadrant.

Question 3

Consider the points $R(3, 0)$, $A(0, -2)$, $M(-5, -2)$ and $P(-5, 2)$. If they are joined in the same order, predict:

Solution. Before drawing anything, look for shared coordinates — that is where parallel and perpendicular sides hide.

$$A(0, -2) \text{ and } M(-5, -2) \text{ share } y = -2 \quad\Rightarrow\quad AM \text{ is horizontal}$$ $$M(-5, -2) \text{ and } P(-5, 2) \text{ share } x = -5 \quad\Rightarrow\quad MP \text{ is vertical}$$

(i) Two sides of RAMP that are perpendicular to each other

A horizontal line and a vertical line always meet at a right angle, and $AM$ and $MP$ meet at the shared vertex $M$.

(ii) One side of RAMP that is parallel to one of the axes

There are in fact two, and it is worth naming both.

(iii) Two points that are mirror images of each other in one axis. Which axis will this be?

Mirror images in the $x$-axis have the same $x$-coordinate and opposite $y$-coordinates. Scanning the four points, $M(-5, -2)$ and $P(-5, 2)$ fit exactly: same $x$, and $-2$ and $2$ are opposites.

Plotting the four points confirms all three predictions: $RAMP$ comes out as a quadrilateral with a right angle at $M$, a horizontal base $AM$ and a vertical side $MP$.

Question 4

Plot point $Z(5, -6)$ on the Cartesian plane. Construct a right-angled triangle $IZN$ and find the lengths of the three sides. (Answers may differ from person to person.)

Solution. $Z(5, -6)$ lies in Quadrant IV. The easiest right-angled triangle to build on a given point is one with the right angle at that point and its two legs along the grid: move a whole number of units vertically for one leg, and horizontally for the other.

Take $I$ four units above $Z$ and $N$ three units to the right of $Z$:

$$I(5,\ -2), \qquad Z(5,\ -6), \qquad N(8,\ -6)$$

$ZI$ is vertical and $ZN$ horizontal, so the angle at $Z$ is a right angle. The legs are single subtractions, and the hypotenuse comes from the Baudhāyana–Pythagoras theorem:

$$ZI = -2 – (-6) = 4, \qquad ZN = 8 – 5 = 3$$ $$IN = \sqrt{(8-5)^2 + \bigl(-6 – (-2)\bigr)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

Question 5

What would a system of coordinates be like if we did not have negative numbers? Would this system allow us to locate all the points on a $2$-D plane?

Solution. Every coordinate would have to be zero or positive. Since $x \ge 0$ places a point on or right of the $y$-axis, and $y \ge 0$ places it on or above the $x$-axis, the only points that could be named are those in Quadrant I together with the two positive half-axes and the origin.

That is a quarter of the plane. Points such as $(-5, 3)$, $(-2, -7)$ and $(4, -6)$ — everything in Quadrants II, III and IV — would have no name at all.

It is worth knowing that this is not a hypothetical. When coordinates were first used, only the positive quarter was drawn; negative coordinates came later, and they are precisely what turned a useful sketching device into a system that covers the whole plane. You could patch the gap by adding words — “$5$ left, $3$ up” — but then the neatness is gone: the rules for distance and midpoint work as arithmetic only because a direction can be carried by a sign.

Question 6

*Are the points $M(-3, -4)$, $A(0, 0)$ and $G(6, 8)$ on the same straight line? Suggest a method to check this without plotting and joining the points.

Solution. Two methods work, and both avoid drawing.

Method 1 — compare the steps. Going from $M$ to $A$, $x$ increases by $3$ and $y$ by $4$. Going from $A$ to $G$, $x$ increases by $6$ and $y$ by $8$:

$$M \to A: \; (3,\ 4) \qquad A \to G: \; (6,\ 8) = 2 \times (3,\ 4)$$

The second step is the first step doubled — the same direction, twice as far. Points that move in the same direction from a common point lie on one line.

Method 2 — compare the distances. Collinear points are the only ones for which the two short distances add exactly to the long one:

$$MA = \sqrt{3^2 + 4^2} = 5, \qquad AG = \sqrt{6^2 + 8^2} = 10, \qquad MG = \sqrt{9^2 + 12^2} = \sqrt{225} = 15$$

$$MA + AG = 5 + 10 = 15 = MG$$

The equality is exact, so $A$ lies on the segment $MG$.

Question 7

*Use your method (from Problem 6) to check if the points $R(-5, -1)$, $B(-2, -5)$ and $C(4, -12)$ are on the same straight line. Now plot both sets of points and check your answers.

Solution. Apply the step test.

$$R \to B: \; \bigl(-2 – (-5),\ -5 – (-1)\bigr) = (3,\ -4)$$ $$B \to C: \; \bigl(4 – (-2),\ -12 – (-5)\bigr) = (6,\ -7)$$

If the points were collinear, $(6, -7)$ would have to be a multiple of $(3, -4)$. Doubling the first step gives $(6, -8)$, not $(6, -7)$. The $x$-parts match and the $y$-parts do not, so the two steps point in slightly different directions.

The distance test says the same thing, and shows just how slight the difference is:

$$RB = \sqrt{3^2 + 4^2} = 5, \qquad BC = \sqrt{6^2 + 7^2} = \sqrt{85} \approx 9.2195$$ $$RC = \sqrt{9^2 + 11^2} = \sqrt{202} \approx 14.2127$$ $$RB + BC \approx 14.2195 \quad\text{against}\quad RC \approx 14.2127$$

They differ by about $0.007$ units — roughly the thickness of a pencil line. That is why the question tells you to plot both sets afterwards: on squared paper these three points look collinear, and the eye cannot separate $14.2195$ from $14.2127$. The arithmetic can.

Question 8

*Using the origin as one vertex, plot the vertices of:

Solution. In both parts the origin is fixed and the other two vertices are yours to choose. Choose them on the grid lines, so that the lengths are whole numbers you can check.

(i) A right-angled isosceles triangle

Put the two equal sides along the axes, so the right angle sits at the origin and both legs are easy to measure:

$$O(0,\ 0), \qquad (4,\ 0), \qquad (0,\ 4)$$

$OA$ lies along the $x$-axis and $OB$ along the $y$-axis, so the angle at $O$ is a right angle; both legs have length $4$, so the triangle is isosceles. The hypotenuse is $\sqrt{4^2 + 4^2} = 4\sqrt{2}$.

(ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV

Quadrant III needs both coordinates negative; Quadrant IV needs $x > 0$ and $y < 0$. Choosing the two points as mirror images in the $y$-axis makes them automatically equidistant from the origin:

$$O(0,\ 0), \qquad (-3,\ -4) \text{ in Quadrant III}, \qquad (3,\ -4) \text{ in Quadrant IV}$$

$$\text{both sides from } O = \sqrt{3^2 + 4^2} = 5$$

Two sides equal, so the triangle is isosceles; the third side runs from $(-3, -4)$ to $(3, -4)$ and has length $6$.

Question 9

*The following table shows the coordinates of points $S$, $M$ and $T$. In each case, state whether $M$ is the midpoint of segment $ST$. Justify your answer. When $M$ is the mid-point of $ST$, can you find any connection between the coordinates of $M$, $S$ and $T$?

Solution. Work out the true midpoint of $ST$ in each row and compare it with the $M$ that is offered. The midpoint of a segment has each coordinate equal to the average of the corresponding coordinates of the endpoints.

Row 1: $S(-3, 0)$, $T(3, 0)$.

$$\left( \frac{-3 + 3}{2},\ \frac{0 + 0}{2} \right) = (0,\ 0)$$

which is the given $M$. Yes.

Row 2: $S(2, 3)$, $T(4, 5)$.

$$\left( \frac{2 + 4}{2},\ \frac{3 + 5}{2} \right) = (3,\ 4)$$

which is the given $M$. Yes.

Row 3: $S(0, 0)$, $T(0, -10)$.

$$\left( \frac{0 + 0}{2},\ \frac{0 + (-10)}{2} \right) = (0,\ -5)$$

but $M$ is given as $(0, 5)$. No — the midpoint is five units below the origin, not above it. $M(0,5)$ is not even between $S$ and $T$.

Row 4: $S(-8, 7)$, $T(6, -3)$.

$$\left( \frac{-8 + 6}{2},\ \frac{7 + (-3)}{2} \right) = (-1,\ 2)$$

but $M$ is given as $(0, -2)$. No — neither coordinate matches.

Question 10

*Use the connection you found to find the coordinates of $B$ given that $M(-7, 1)$ is the midpoint of $A(3, -4)$ and $B(x, y)$.

Solution. Write down the midpoint relation and solve it for the unknown endpoint. From $M = \left( \frac{x_A + x_B}{2}, \frac{y_A + y_B}{2} \right)$:

$$\frac{3 + x}{2} = -7 \quad\Rightarrow\quad 3 + x = -14 \quad\Rightarrow\quad x = -17$$ $$\frac{-4 + y}{2} = 1 \quad\Rightarrow\quad -4 + y = 2 \quad\Rightarrow\quad y = 6$$

It is worth reading the rearranged form as a rule in its own right: the far endpoint is $B = 2M – A$, that is, twice the midpoint minus the known endpoint.

Check: the midpoint of $A(3, -4)$ and $B(-17, 6)$ is $\left( \frac{3 – 17}{2}, \frac{-4 + 6}{2} \right) = (-7, 1)$ — the given $M$.

Question 11

*Let $P$, $Q$ be points of trisection of $AB$, with $P$ closer to $A$ and $Q$ closer to $B$. Using your knowledge of how to find the coordinates of the midpoint of a segment, how would you find the coordinates of $P$ and $Q$? Do this for the case when the points are $A(4, 7)$ and $B(16, -2)$.

Solution. The midpoint formula averages the two endpoints equally. Trisection points are not halfway, so they need unequal weights — but the idea is the same, and there is a neat way to see it using midpoints alone.

$P$, $Q$ cut $AB$ into three equal pieces, so $AP = PQ = QB$. That means:

• $P$ is the midpoint of $A$ and $Q$, because $AP = PQ$;
• $Q$ is the midpoint of $P$ and $B$, because $PQ = QB$.

Each trisection point is therefore the midpoint of its neighbours. To get them directly, take one third of the journey from $A$ to $B$ for $P$, and two thirds for $Q$:

$$B – A = (16 – 4,\ -2 – 7) = (12,\ -9)$$

$$P = A + \tfrac{1}{3}(12,\ -9) = (4 + 4,\ 7 – 3) = (8,\ 4)$$ $$Q = A + \tfrac{2}{3}(12,\ -9) = (4 + 8,\ 7 – 6) = (12,\ 1)$$

Both checks work: the midpoint of $P(8,4)$ and $B(16,-2)$ is $\left( \frac{8+16}{2}, \frac{4-2}{2} \right) = (12, 1) = Q$, and the midpoint of $A(4,7)$ and $Q(12,1)$ is $\left( \frac{4+12}{2}, \frac{7+1}{2} \right) = (8, 4) = P$.

Question 12

*Given the points $A(1, -8)$, $B(-4, 7)$ and $C(-7, -4)$:

Solution. A point lies on a circle centred at the origin exactly when its distance from the origin equals the radius. Since every distance here is a square root, compare the squares — it avoids surds entirely.

(i) Show that they lie on a circle $K$ whose centre is the origin $O(0, 0)$. What is the radius of circle $K$?

$$OA^2 = 1^2 + (-8)^2 = 1 + 64 = 65$$ $$OB^2 = (-4)^2 + 7^2 = 16 + 49 = 65$$ $$OC^2 = (-7)^2 + (-4)^2 = 49 + 16 = 65$$

All three are the same number, so all three points are the same distance from the origin — which is what it means to lie on one circle centred there.

(ii) Given the points $D(-5, 6)$ and $E(0, 9)$, check whether $D$ and $E$ lie within the circle, on the circle, or outside the circle $K$.

Compare each squared distance with $65$:

$$OD^2 = (-5)^2 + 6^2 = 25 + 36 = 61 \; < \; 65$$ $$OE^2 = 0^2 + 9^2 = 81 \; > \; 65$$

Question 13

*The midpoints of the sides of triangle $ABC$ are the points $D$, $E$ and $F$. Given that the coordinates of $D$, $E$ and $F$ are $(5, 1)$, $(6, 5)$ and $(0, 3)$ respectively, find the coordinates of $A$, $B$ and $C$.

Solution. Take the usual labelling: $D$ is the midpoint of $BC$, $E$ of $CA$, and $F$ of $AB$. Then

$$B + C = 2D, \qquad C + A = 2E, \qquad A + B = 2F$$

where each equation is really two — one for the $x$-coordinates, one for the $y$. Adding all three gives $2(A + B + C) = 2(D + E + F)$, so

$$A + B + C = D + E + F$$

Subtract the first equation from this to isolate $A$, and similarly for the others:

$$A = (D + E + F) – (B + C) = E + F – D$$ $$B = F + D – E, \qquad C = D + E – F$$

Now substitute, coordinate by coordinate:

$$A = (6 + 0 – 5,\ 5 + 3 – 1) = (1,\ 7)$$ $$B = (0 + 5 – 6,\ 3 + 1 – 5) = (-1,\ -1)$$ $$C = (5 + 6 – 0,\ 1 + 5 – 3) = (11,\ 3)$$

Check by going back the other way: the midpoint of $BC$ is $\left( \frac{-1+11}{2}, \frac{-1+3}{2} \right) = (5, 1) = D$; of $CA$, $\left( \frac{11+1}{2}, \frac{3+7}{2} \right) = (6, 5) = E$; of $AB$, $\left( \frac{1-1}{2}, \frac{7-1}{2} \right) = (0, 3) = F$. All three match.

Question 14

A city has two main roads which cross each other at the centre of the city. These two roads are along the North–South (N–S) direction and East–West (E–W) direction. All the other streets of the city run parallel to these roads and are $200$ m apart. There are $10$ streets in each direction.

Solution. The model is a square grid. Streets running N–S are vertical lines, streets running E–W are horizontal lines, and consecutive lines in each family are $200$ m apart.

(i) Using $1$ cm $= 200$ m, draw a model of the city in your notebook. Represent the roads/streets by single lines.

At that scale $200$ m becomes $1$ cm, so consecutive streets are drawn $1$ cm apart. Ten vertical lines at $1$ cm spacing span $9$ cm, and likewise for the ten horizontal lines.

(ii) Find (a) how many street intersections can be referred to as $(4, 3)$, and (b) how many can be referred to as $(3, 4)$.

The convention is the one given in the question: the first number names a street running N–S, the second names a street running E–W. Each family is numbered once, $1$ to $10$, so “the 4th N–S street” picks out exactly one vertical line and “the 3rd E–W street” picks out exactly one horizontal line.

Two lines that are not parallel meet at exactly one point. So each label names exactly one crossing — never two, and never none.

$$\text{(a) } (4, 3): \text{ one crossing} \qquad \text{(b) } (3, 4): \text{ one crossing}$$

And they are not the same crossing. $(4, 3)$ is on the 4th vertical and the 3rd horizontal; $(3, 4)$ is on the 3rd vertical and the 4th horizontal. Swapping the numbers moves you to a different corner of the grid — which is precisely why coordinates are written as ordered pairs.

Question 15

A computer graphics program displays images on a rectangular screen whose coordinate system has the origin at the bottom-left corner. The screen is $800$ pixels wide and $600$ pixels high. A circular icon of radius $80$ pixels is drawn with its centre at the point $A(100, 150)$. Another circular icon of radius $100$ pixels is drawn with its centre at the point $B(250, 230)$. Determine:

Solution. With the origin at the bottom-left corner, the screen is the region $0 \le x \le 800$, $0 \le y \le 600$.

(i) Whether any part of either circle lies outside the screen

A circle of radius $r$ centred at $(h, k)$ reaches from $h – r$ to $h + r$ horizontally and from $k – r$ to $k + r$ vertically. It is wholly on screen exactly when all four of those numbers lie inside the screen’s range.

First icon, centre $A(100, 150)$, $r = 80$:

$$x: \; 100 \pm 80 \;\Rightarrow\; 20 \text{ to } 180 \qquad y: \; 150 \pm 80 \;\Rightarrow\; 70 \text{ to } 230$$

Second icon, centre $B(250, 230)$, $r = 100$:

$$x: \; 250 \pm 100 \;\Rightarrow\; 150 \text{ to } 350 \qquad y: \; 230 \pm 100 \;\Rightarrow\; 130 \text{ to } 330$$

Every one of these lies between $0$ and $800$ horizontally and between $0$ and $600$ vertically. The tightest margin is the first icon’s left edge, at $x = 20$ — twenty pixels clear of the screen edge.

(ii) Whether the two circles intersect each other

Two circles meet when the distance between their centres is less than the sum of the radii but more than the difference — close enough to reach each other, far enough apart that neither swallows the other.

$$AB = \sqrt{(250 – 100)^2 + (230 – 150)^2} = \sqrt{150^2 + 80^2} = \sqrt{22500 + 6400} = \sqrt{28900} = 170$$

$$r_1 + r_2 = 80 + 100 = 180, \qquad r_2 – r_1 = 100 – 80 = 20$$

$$20 \; < \; 170 \; < \; 180$$

Question 16

Plot the points $A(2, 1)$, $B(-1, 2)$, $C(-2, -1)$ and $D(1, -2)$ in the coordinate plane. Is $ABCD$ a square? Can you explain why? What is the area of this square?

Solution. None of the sides is parallel to an axis here, so every length has to come from the distance formula. Take the four sides in order:

$$AB = \sqrt{(-1-2)^2 + (2-1)^2} = \sqrt{9 + 1} = \sqrt{10}$$ $$BC = \sqrt{(-2+1)^2 + (-1-2)^2} = \sqrt{1 + 9} = \sqrt{10}$$ $$CD = \sqrt{(1+2)^2 + (-2+1)^2} = \sqrt{9 + 1} = \sqrt{10}$$ $$DA = \sqrt{(2-1)^2 + (1+2)^2} = \sqrt{1 + 9} = \sqrt{10}$$

All four sides are equal, so $ABCD$ is at least a rhombus. Equal sides alone do not make a square, though — a rhombus can lean. The test that settles it is the diagonals:

$$AC = \sqrt{(-2-2)^2 + (-1-1)^2} = \sqrt{16 + 4} = \sqrt{20}$$ $$BD = \sqrt{(1+1)^2 + (-2-2)^2} = \sqrt{4 + 16} = \sqrt{20}$$

A quadrilateral with four equal sides and equal diagonals is a square. (A leaning rhombus has one long diagonal and one short.)

The area is the side squared:

$$\text{area} = \left( \sqrt{10} \right)^2 = 10 \text{ square units}$$

As a check, the area of a square is also half the square of a diagonal: $\frac{1}{2} \times 20 = 10$. The two agree.

Common mistakes

• Question 2, giving a single point for $H$. The line through $W$ parallel to the $y$-axis contains infinitely many points. Only the $x$-coordinate is determined, and the answer has to say so — $(-5, y)$ — rather than pick a convenient $y$ and present it as the answer.
• Question 2, forgetting the third possibility. Listing Quadrants II and III is right as far as it goes, but $(-5, 0)$ lies on the negative $x$-axis, and points on an axis are in no quadrant at all.
• Question 7, trusting the plot. These three points miss collinearity by about $0.007$ units, which no drawing can show. The question deliberately asks you to plot after calculating, and a student who plots first will almost certainly answer “yes”. Whenever a collinearity question comes with an instruction to check by plotting, expect the plot to be the trap rather than the test.
• Question 9, Row 3, comparing only the $x$-coordinates. Both $M$ and the true midpoint have $x = 0$, so a quick glance says “matches”. The $y$-coordinates are $5$ and $-5$ — the given $M$ is on the opposite side of the origin, and is not even between $S$ and $T$.
• Question 10, averaging instead of solving. Treating $M$ as an endpoint and computing $\left( \frac{3 + (-7)}{2}, \frac{-4 + 1}{2} \right) = (-2, -1.5)$. $M$ is the midpoint, so it is the answer to an averaging, not an input to one: $B = 2M – A$.
• Question 11, bisecting twice. Halving $AB$ and then halving again gives the points a quarter and a half of the way along — not thirds. Trisection needs weights of $\frac13$ and $\frac23$; the midpoint idea enters only as the observation that each trisection point is the midpoint of its two neighbours.
• Question 12(ii), comparing $61$ and $81$ with $\sqrt{65}$. Squared distances must be compared with the radius squared. Testing $61$ against $8.06$ makes an interior point look exterior.
• Question 14, answering “many” or “none”. Each number in the pair selects one street, and two non-parallel streets cross exactly once — so the answer is one, both times. The question asks it twice to make a different point: $(4,3)$ and $(3,4)$ are different crossings, because a coordinate pair is ordered.
• Question 15(ii), comparing the centre distance with one radius. Since $170 > 100$, it is tempting to say $A$’s centre is beyond $B$’s circle so they cannot meet. Two circles overlap when $|r_2 – r_1| < AB < r_1 + r_2$; here $170 < 180$, so they do.
• Question 16, stopping at four equal sides. That proves a rhombus, not a square. A rhombus with a $60^\circ$ corner also has four equal sides. The diagonals have to be checked — or one right angle proved — before the word “square” is earned.

Practise next

• Chapter 2, Introduction to Linear Polynomials — the next chapter, where the same plane is used to draw the graph of a relationship rather than to plot isolated points.
• Exercise Set 1.2, Question 3 — worth revisiting after Question 16 here. Both ask what shape a set of coordinates makes; there the sides were parallel to the axes and could be read off, here they had to be computed. Seeing the two side by side is what fixes when the distance formula is actually needed.