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

Book page 12–14 Updated on2026-09-08

Q1.

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

Answer

The two axes meet at the origin.

x-coordinate = 0, y-coordinate = 0
The point is O (0, 0)

Why it happens: The point of intersection lies on the y-axis, so its perpendicular distance from the y-axis is 0, giving x = 0. It also lies on the x-axis, so y = 0. And these two conditions together pin down a single point: only (0, 0) satisfies both. That is why two intersecting lines can share exactly one point.

Q2.

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?

Answer

H = (−5, k), where k can be any real number

Why the x-coordinate is forced: A line parallel to the y-axis is a vertical line. Every point on it is the same perpendicular distance from the y-axis and on the same side of it, so every point on it has the same x-coordinate. W has x = −5, so the whole line is x = −5, and H must share that value. Its y-coordinate, however, is completely free — moving up or down the line changes nothing about the distance from the y-axis.

Quadrants: since x = −5 is negative, H lies to the left of the y-axis. So:

Value of k	Point	Where it lies
k > 0	(−5, k)	Quadrant II
k = 0	(−5, 0)	on the x-axis, in no quadrant
k < 0	(−5, k)	Quadrant III

So H can lie only in Quadrant II or Quadrant III — never in I or IV, because those need a positive x-coordinate.

Tip: x = −5 is the equation of that vertical line. In Grade 10 you will meet such equations everywhere; the idea starts right here, with one coordinate held fixed while the other roams.

Q3.

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: (i) Two sides of RAMP that are perpendicular to each other. (ii) One side of RAMP that is parallel to one of the axes. (iii) Two points that are mirror images of each other in one axis. Which axis will this be? Now plot the points and verify your predictions.

Answer

Predict first, by reading the coordinates only.

(i) AM and MP are perpendicular.

A (0, −2) and M (−5, −2) share y = −2 → AM is horizontal
M (−5, −2) and P (−5, 2) share x = −5 → MP is vertical
Horizontal ⊥ vertical → AM ⊥ MP, the right angle being at M

(ii) AM is parallel to the x-axis (both ends have y = −2). MP is parallel to the y-axis, so either answer is acceptable; the other two sides, RA and PR, are parallel to neither.

(iii) M (−5, −2) and P (−5, 2) are mirror images in the x-axis.

Same x-coordinate, opposite y-coordinates: −2 and +2
Reflection in the x-axis sends (x, y) to (x, −y) → M ↔ P ✓

Quadrilateral RAMP. AM is horizontal, MP is vertical, so the angle at M is a right angle; M and P are mirror images in the x-axis.

Verification by calculation:

RA = √((0 − 3)² + (−2 − 0)²) = √(9 + 4) = √13
AM = |−5 − 0| = 5
MP = |2 − (−2)| = 4
PR = √((3 − (−5))² + (0 − 2)²) = √(64 + 4) = √68 = 2√17

Right angle at M: the shift M→A is (5, 0), the shift M→P is (0, 4). One is purely horizontal and the other purely vertical, so they meet at 90°.

Why the predictions could be made without a diagram: A shared y-coordinate is precisely the condition for a segment to be horizontal; a shared x-coordinate is the condition for it to be vertical. Equal x with opposite y is precisely reflection in the x-axis. Each geometric property has an exact arithmetic signature — that is what coordinate geometry buys you.

Did you know? RAMP is not a rectangle, though it has a right angle: RA = √13 and PR = 2√17 are unequal and neither is parallel to an axis. One right angle is not enough.

Q4.

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

Answer

Z (5, −6) is in Quadrant IV: 5 to the right of O, then 6 down. The easiest way to guarantee a right angle at Z is to take one other vertex directly above it and one directly beside it.

Choose I = (5, −2) — same x as Z, so ZI is vertical
Choose N = (8, −6) — same y as Z, so ZN is horizontal
→ ∡IZN = 90°

A right-angled triangle IZN with the right angle at Z. The legs are chosen parallel to the axes, so their lengths are simple differences.

ZI = |−2 − (−6)| = 4 units
ZN = |8 − 5| = 3 units
IN = √((8 − 5)² + (−6 − (−2))²) = √(9 + 16) = √25 = 5 units

Check: ZN² + ZI² = 9 + 16 = 25 = IN² ✓

Why this construction always works: Making one leg vertical and the other horizontal guarantees the right angle, because the two axis directions are perpendicular by definition. It also makes both leg lengths plain subtractions, so only the hypotenuse needs the theorem. Deliberately choosing legs of 3 and 4 turns the hypotenuse into a whole number, 5.

Try This: Your I and N will very likely differ from these — the question says so. Test your own triangle the same way: verify that the sum of the squares of the two shorter sides equals the square of the longest. If it does not, your triangle is not right-angled.

Q5.

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?

Answer

No. Without negative numbers you could label only one quarter of the plane.

Only x ≥ 0 and y ≥ 0 could be written → only Quadrant I survives
The axes shrink from full lines to two rays leaving O
Quadrants II, III and IV would have no addresses at all

Why it fails: Each coordinate has to record two things about a shift — how far, and in which of the two opposite directions. A number without a sign records only the size. So (3, 2) and (−3, 2) would collapse to the same label “3, 2”, and the same address would name two different points. The correspondence between points and pairs of numbers would stop being one-to-one, and that correspondence is the whole idea of the coordinate system. Signs are what let a single number carry a direction as well as a magnitude.

You could patch it by writing “3 left, 2 up” in words — but then “left” is doing exactly the job of the minus sign, and the neat algebra is lost. The distance formula, for instance, works because x₂ − x₁ is a signed number that gets squared.

Did you know? This is the connection the chapter opens with. Brahmagupta (c. 628 CE) was the first to treat zero and negative numbers as genuine algebraic quantities. Without his work, the four-quadrant plane you are drawing in this chapter could not exist — Descartes himself first worked mostly in what we would call the first quadrant.

Q6.

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.

Answer

Yes, M, A and G are collinear, with A lying between M and G.

Method 1 — compare the shifts. Read off how far you move horizontally and vertically along each step:

M → A: shift = (0 − (−3), 0 − (−4)) = (3, 4)
A → G: shift = (6 − 0, 8 − 0) = (6, 8) = 2 × (3, 4)
Vertical shift ÷ horizontal shift: 4/3 for the first step, 8/6 = 4/3 for the second → equal
→ the direction never changes → the three points lie on one line

Method 2 — use distances. Three points are collinear exactly when the longest of the three distances equals the sum of the other two:

MA = √(3² + 4²) = √25 = 5
AG = √(6² + 8²) = √100 = 10
MG = √((6 − (−3))² + (8 − (−4))²) = √(81 + 144) = √225 = 15
MA + AG = 5 + 10 = 15 = MG ✓ → collinear

Why the distance test works: If M, A, G were the vertices of a genuine triangle, the triangle inequality would give MA + AG > MG strictly. Equality can happen only when the triangle has collapsed — that is, when A lies on the segment MG. So checking MA + AG = MG is a complete test, and it needs no drawing.

Why the shift test works: Walking along a straight line, the ratio of vertical rise to horizontal run is fixed — that fixed ratio is what “straight” means in coordinates. If the ratio changes between one step and the next, the path has turned. Method 1 is usually quicker because it avoids square roots altogether, and it also handles the case where the middle point is not between the other two.

Tip: With Method 1, cross-multiply instead of dividing, to keep everything in whole numbers: 4 × 6 = 24 and 3 × 8 = 24, equal, so collinear. This also avoids trouble when a horizontal shift is 0.

Q7.

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.

Answer

No — R, B and C are not collinear, although they come remarkably close.

R → B: shift = (−2 − (−5), −5 − (−1)) = (3, −4)
B → C: shift = (4 − (−2), −12 − (−5)) = (6, −7)

Cross-multiply to compare 3 : −4 with 6 : −7 —
3 × (−7) = −21 and (−4) × 6 = −24
−21 ≠ −24 → the direction changes at B → not collinear

The distance test agrees, but only just:

RB = √(3² + 4²) = √25 = 5
BC = √(6² + 7²) = √85 ≈ 9.2195
RC = √(9² + 11²) = √(81 + 121) = √202 ≈ 14.2127
RB + BC ≈ 14.2195, while RC ≈ 14.2127
Difference ≈ 0.007 units — small, but not zero → not collinear

Why plotting cannot settle this one: At the scale 1 cm = 1 unit, the bend at B is about 0.07 mm — thinner than your pencil line. Draw the three points and they will look exactly like a straight line. That is the real lesson of this pair of questions: in Q6 the algebra confirmed what a drawing would suggest, but here a drawing would have misled you. An exact test beats a picture.

Check it yourself: If C had been (4, −13) instead, the second shift would be (6, −8) = 2 × (3, −4) and the three points would be collinear. One unit of difference in a single coordinate decides the matter — and the eye cannot see it.

Q8.

Using the origin as one vertex, plot the vertices of: (i) A right-angled isosceles triangle. (ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.

Answer

Answers may differ. Two convenient choices:

Two triangles with a vertex at O. In blue, a right-angled isosceles triangle with its legs on the axes; in green, an isosceles triangle straddling Quadrants III and IV.

(i) Right-angled isosceles triangle: O (0, 0), (4, 0), (0, 4).

One leg on the x-axis: length |4 − 0| = 4
One leg on the y-axis: length |4 − 0| = 4 → equal → isosceles
The axes are perpendicular → the angle at O is 90° → right-angled
Hypotenuse = √(4² + 4²) = √32 = 4√2 ≈ 5.66 units

(ii) Isosceles triangle with one vertex in Quadrant III and one in Quadrant IV: O (0, 0), U (−3, −4), V (3, −4).

U (−3, −4) has signs (−, −) → Quadrant III ✓
V (3, −4) has signs (+, −) → Quadrant IV ✓
OU = √((−3)² + (−4)²) = √25 = 5
OV = √(3² + (−4)²) = √25 = 5 → OU = OV → isosceles
UV = |3 − (−3)| = 6

Why picking U and V as a −/+ pair guarantees the isosceles property: U and V have opposite x-coordinates and the same y-coordinate, so they are mirror images of each other in the y-axis. The origin lies on that mirror line, so it is equidistant from any pair of mirror-image points. You get OU = OV for free, without computing either — the symmetry does the proof.

Try This: Try to make (ii) also right-angled at O. You need OU ⊥ OV, and with U (−a, −b), V (a, −b) that happens only when a = b — for example O, (−3, −3), (3, −3). Check that the two legs are then √18 each and the base is 6, and that 18 + 18 = 36 ✓.

Q9.

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. [S, M, T: (−3, 0), (0, 0), (3, 0) | (2, 3), (3, 4), (4, 5) | (0, 0), (0, 5), (0, −10) | (−8, 7), (0, −2), (6, −3)] When M is the mid-point of ST, can you find any connection between the coordinates of M, S and T?

Answer

Test each row by averaging the coordinates of S and T and comparing with M.

S	M	T	Is M the midpoint of ST?	Reason
(−3, 0)	(0, 0)	(3, 0)	Yes	((−3 + 3)/2, (0 + 0)/2) = (0, 0) = M
(2, 3)	(3, 4)	(4, 5)	Yes	((2 + 4)/2, (3 + 5)/2) = (3, 4) = M
(0, 0)	(0, 5)	(0, −10)	No	((0 + 0)/2, (0 + (−10))/2) = (0, −5), not (0, 5)
(−8, 7)	(0, −2)	(6, −3)	No	((−8 + 6)/2, (7 + (−3))/2) = (−1, 2), not (0, −2)

The connection: each coordinate of M is the average of the corresponding coordinates of S and T.

If M is the midpoint of ST, then
x_M = (x_S + x_T)/2 and y_M = (y_S + y_T)/2

Why the averaging rule is true: M is the midpoint exactly when the journey S → M repeats itself as M → T. Compare the two journeys one coordinate at a time. The horizontal shifts must be equal, so x_M − x_S = x_T − x_M, which rearranges to 2x_M = x_S + x_T. The same argument on the vertical shifts gives 2y_M = y_S + y_T. Since each coordinate can be handled separately, the midpoint of a slanted segment is no harder than the midpoint of a horizontal one.

Check it yourself: Equal distances alone are not enough. In row 3, M (0, 5) is 5 units from S and 15 units from T — it is not even on the segment, which runs downwards from (0, 0) to (0, −10). A midpoint must lie on the segment, and the averaging rule guarantees that automatically.

Q10.

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).

Answer

Apply the averaging rule one coordinate at a time and solve for the unknown.

(3 + x)/2 = −7 → 3 + x = −14 → x = −17
(−4 + y)/2 = 1 → −4 + y = 2 → y = 6

B = (−17, 6)

Check: midpoint of A (3, −4) and B (−17, 6) = ((3 − 17)/2, (−4 + 6)/2) = (−7, 1) = M ✓

Why the answer is so far from A: M is not between the numbers 3 and −7 by a little — going from A to M is a shift of (−10, 5), and B must be a second, identical shift beyond M: (−7 − 10, 1 + 5) = (−17, 6). Thinking in shifts gives the answer in one line and is a useful check on the algebra.

Tip: A pair of coordinates gives two independent equations, one for x and one for y. They never mix, so you always solve them separately — never as a simultaneous pair.

Q11.

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).

Answer

The method. The midpoint rule works because you take half of the shift from A to B. Trisection points need one-third and two-thirds of the same shift.

Shift from A to B = (x_B − x_A, y_B − y_A)
Midpoint = A + ½ of the shift
P = A + ⅓ of the shift → P = ((2x_A + x_B)/3, (2y_A + y_B)/3)
Q = A + ⅔ of the shift → Q = ((x_A + 2x_B)/3, (y_A + 2y_B)/3)

For A (4, 7) and B (16, −2):

Shift from A to B = (16 − 4, −2 − 7) = (12, −9)
One-third of the shift = (4, −3)

P = (4 + 4, 7 − 3) = (8, 4)
Q = (8 + 4, 4 − 3) = (12, 1)
and B = (12 + 4, 1 − 3) = (16, −2) ✓

Verification using midpoints only, as the question asks:

Midpoint of A (4, 7) and Q (12, 1) = ((4 + 12)/2, (7 + 1)/2) = (8, 4) = P ✓
Midpoint of P (8, 4) and B (16, −2) = ((8 + 16)/2, (4 − 2)/2) = (12, 1) = Q ✓

Why the midpoint idea extends: The midpoint formula is really a statement about splitting each shift in a fixed ratio, and the coordinates handle that splitting independently. If P cuts AB in the ratio 1 : 2, then the horizontal shift 12 is cut as 4 and 8, and the vertical shift −9 as −3 and −6 — the same ratio in both. That is why a single fraction of the shift moves you to the right point.

Check it yourself: AP = √(4² + 3²) = 5, PQ = √(4² + 3²) = 5, QB = √(4² + 3²) = 5, and AB = √(12² + 9²) = √225 = 15 = 3 × 5. The three pieces really are equal, so P and Q are genuine points of trisection.

Q12.

(i) Given the points A (1, − 8), B (− 4, 7) and C (−7, − 4), show that they lie on a circle K whose center is the origin O (0, 0). What is the radius of circle K? (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.

Answer

(i) A circle with centre O is the set of all points at one fixed distance from O. So compute OA, OB, OC and see whether they agree.

OA = √(1² + (−8)²) = √(1 + 64) = √65
OB = √((−4)² + 7²) = √(16 + 49) = √65
OC = √((−7)² + (−4)²) = √(49 + 16) = √65

OA = OB = OC = √65 → all three lie on one circle centred at O
Radius of K = √65 ≈ 8.06 units

Why this is a complete proof: “Lies on the circle with centre O and radius r” means exactly “is at distance r from O” — that is the definition of a circle, not a property of it. So showing the three distances are equal is showing the three points are concyclic about O. Notice also why the three answers came out equal: 1² + 8², 4² + 7² and 7² + 4² all total 65, even though the points sit in three different quadrants. Squaring wipes out the signs.

(ii) Compare each distance with the radius. It is cleaner to compare the squares and avoid roots altogether:

r² = 65

OD² = (−5)² + 6² = 25 + 36 = 61 → 61 < 65 → D lies inside the circle
OE² = 0² + 9² = 81 → 81 > 65 → E lies outside the circle

In lengths: OD = √61 ≈ 7.81 < 8.06, and OE = 9 > 8.06.

Why comparing squares is safe: Distances are never negative, and squaring is increasing on non-negative numbers. So OD < r exactly when OD² < r². Working with 61, 65 and 81 keeps everything in whole numbers and removes any doubt caused by rounding — √61 ≈ 7.81 and √65 ≈ 8.06 are close enough that a careless decimal could mislead you.

Tip: This gives you a three-way test for any point P and any circle of centre C, radius r: compare CP² with r² — less means inside, equal means on, greater means outside.

Q13.

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.

Answer

Take D as the midpoint of BC, E as the midpoint of CA and F as the midpoint of AB.

Midpoint rule, written as sums:
B + C = 2D = (10, 2)  …(1)
C + A = 2E = (12, 10)  …(2)
A + B = 2F = (0, 6)  …(3)

Add all three: 2(A + B + C) = (22, 18) → A + B + C = (11, 9)

A = (A + B + C) − (B + C) = (11, 9) − (10, 2) = (1, 7)
B = (A + B + C) − (C + A) = (11, 9) − (12, 10) = (−1, −1)
C = (A + B + C) − (A + B) = (11, 9) − (0, 6) = (11, 3)

Check all three midpoints:

Midpoint of BC = ((−1 + 11)/2, (−1 + 3)/2) = (5, 1) = D ✓
Midpoint of CA = ((11 + 1)/2, (3 + 7)/2) = (6, 5) = E ✓
Midpoint of AB = ((1 − 1)/2, (7 − 1)/2) = (0, 3) = F ✓

Why adding the three equations is the key step: Each equation involves two of the three unknowns, so no single one can be solved on its own. Adding them makes every vertex appear exactly twice, which produces A + B + C. Subtracting any one of the original equations from that total then isolates a single vertex. This trick — find the total first, then peel off — is worth remembering; it works because each coordinate can be handled separately, so what looks like one vector equation is really two ordinary linear equations.

Did you know? ∆DEF is called the medial triangle of ∆ABC. The three equations always have exactly one solution, so a triangle is completely determined by the midpoints of its sides. Pairing the letters differently (say D with CA instead of BC) only renames A, B and C — the triangle you get is the same one.

Q14.

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. (i) Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roads/streets by single lines. (ii) There are street intersections in the model. Each street intersection is formed by two streets — one running in the N–S direction and another in the E–W direction. Each street intersection is referred to in the following manner: If the second street running in the N–S direction and 5th street in the E–W direction meet at some crossing, then we call this street intersection (2, 5). Using this convention, find: (a) how many street intersections can be referred to as (4, 3). (b) how many street intersections can be referred to as (3, 4).

Answer

(i) Draw 10 vertical lines (the N–S streets) and 10 horizontal lines (the E–W streets), each set spaced 1 cm apart, since 200 m at the scale 1 cm = 200 m is exactly 1 cm. Number the N–S streets 1 to 10 from left to right and the E–W streets 1 to 10 from bottom to top. The grid has 10 × 10 = 100 crossings.

The city model: 10 N–S streets and 10 E–W streets, 1 cm apart. (4, 3) and (3, 4) are two different crossings.

(ii) (a) Exactly one. (b) Exactly one — but a different one.

(4, 3) = crossing of the 4th N–S street with the 3rd E–W street → 1 intersection
(3, 4) = crossing of the 3rd N–S street with the 4th E–W street → 1 intersection
(4, 3) ≠ (3, 4)

Why the count is always one: An N–S street and an E–W street are not parallel, so they meet, and two straight lines that are not parallel meet at exactly one point — they cannot meet twice. Naming one street from each family therefore names one crossing and only one. This is the same reason a point of the plane has only one coordinate pair.

Why (4, 3) and (3, 4) are different places: The convention fixes an order: the first number always names the N–S street, the second always the E–W street. Reading them the other way round sends you 200 m along one road and 200 m back along the other — a different crossing altogether. A postal address works the same way; so does (x, y).

Did you know? The chapter opens with exactly this idea in practice. The Sindhu–Sarasvatī cities laid their streets out N–S and E–W about 10 m apart, so a merchant could reach a warehouse by counting streets in the two directions — a working coordinate system thousands of years before it was written down as algebra.

Q15.

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: (i) whether any part of either circle lies outside the screen. (ii) whether the two circles intersect each other.

Answer

With the origin at the bottom-left corner, the visible screen is exactly the set of points with 0 ≤ x ≤ 800 and 0 ≤ y ≤ 600.

The two icons on the 800 × 600 screen. Both fit inside; their centres are 170 pixels apart.

(i) No — both circles lie entirely on the screen. A circle stays inside the rectangle exactly when the distance from its centre to every edge is at least the radius.

Icon	To left edge	To right edge	To bottom edge	To top edge	Radius	Verdict
A (100, 150)	100	700	150	450	80	smallest gap 80 ≤ 100 → inside
B (250, 230)	250	550	230	370	100	smallest gap 100 ≤ 230 → inside

Icon A spans x from 100 − 80 = 20 to 100 + 80 = 180, and y from 70 to 230 — all within the screen
Icon B spans x from 150 to 350, and y from 130 to 330 — all within the screen

Why checking the four edge distances is enough: The point of a circle nearest to a given straight edge lies on the perpendicular from the centre, at distance (centre-to-edge) − r. If that is still non-negative for all four edges, no part of the circle can cross any edge. Distance to a vertical edge is a difference of x-coordinates only; to a horizontal edge, of y-coordinates only — which is why the screen was set up with its edges along the axes.

(ii) Yes, the two circles intersect — they cut each other at two points.

AB = √((250 − 100)² + (230 − 150)²)
= √(150² + 80²) = √(22500 + 6400) = √28900 = 170 pixels

Sum of radii = 80 + 100 = 180
Difference of radii = 100 − 80 = 20

20 < 170 < 180, that is, |r₁ − r₂| < AB < r₁ + r₂ → the circles meet at two points

Why the two comparisons decide it: If the centres were farther apart than 180 the circles would be completely separate, and if they touched at exactly 180 they would meet at a single point. If the centres were closer than 20 the smaller circle would sit wholly inside the larger one, meeting it nowhere. Only when the centre distance lies strictly between the difference and the sum of the radii can the two boundaries cross — and 170 is in that range, with just 10 pixels to spare before they would separate.

Did you know? Games and drawing programs test for collisions in exactly this way. Comparing AB² = 28900 with (r₁ + r₂)² = 32400 avoids the square root entirely, which is why it is fast enough to run for hundreds of objects in every frame.

Q16.

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?

Answer

Yes, ABCD is a square, and its area is 10 square units.

ABCD with its diagonals. All four sides are √10 and both diagonals are √20 — a tilted square.

The four sides:
AB = √((−1 − 2)² + (2 − 1)²) = √(9 + 1) = √10
BC = √((−2 + 1)² + (−1 − 2)²) = √(1 + 9) = √10
CD = √((1 + 2)² + (−2 + 1)²) = √(9 + 1) = √10
DA = √((2 − 1)² + (1 + 2)²) = √(1 + 9) = √10
→ all four sides equal → ABCD is a rhombus

The two diagonals:
AC = √((−2 − 2)² + (−1 − 1)²) = √(16 + 4) = √20
BD = √((1 + 1)² + (−2 − 2)²) = √(4 + 16) = √20
→ the diagonals are equal → the rhombus is a square

Why equal sides alone are not enough: Four equal sides make a rhombus, and a rhombus can lean over as far as you like without any side changing length. What fixes the angles is the diagonals: in a rhombus the diagonals are equal only when the angles are right angles. You can also see it directly with the converse of the Baudhāyana–Pythagoras Theorem — AB² + BC² = 10 + 10 = 20 = AC², so the angle at B is 90°, and a rhombus with one right angle is a square.

Area, two ways:
side² = (√10)² = 10 square units
½ × d₁ × d₂ = ½ × √20 × √20 = ½ × 20 = 10 square units ✓

Did you know? This square is tilted, so its sides are not whole numbers — yet its area is the whole number 10. That is possible because area only needs the square of the side, and the squaring in the distance formula undoes the square root. The centre of the square is the midpoint of AC, which is ((2 − 2)/2, (1 − 1)/2) = (0, 0) — the square is centred on the origin, which is why the four points came in ± pairs.

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