Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
Reiaan’s room is drawn with a scale of 1 cm : 1 foot. Choose the corner O of the room as the origin and the two walls as axes.
The x-coordinate of every point on the y-axis is 0. The y-coordinate of every point on the x-axis is 0.
The plane with its axes is called the Cartesian plane, coordinate plane or xy-plane. The axes divide it into four quadrants, numbered anticlockwise from the top right.
| Quadrant | Sign of (x, y) | Example |
|---|---|---|
| I | (+, +) | (3, 4) |
| II | (−, +) | (−5, 3) |
| III | (−, −) | (−2, −7) |
| IV | (+, −) | (3, −5) |
If the segment is not parallel to an axis, make a right triangle. Take A (3, 4) and D (7, 1). The horizontal leg is CD = 7 − 3 = 4. The vertical leg is AC = 4 − 1 = 3. By the Baudhāyana–Pythagoras theorem AD = √4² + 3² = 5.
Distance formula: the distance between (x₁, y₁) and (x₂, y₂) is
| Reflection in | (x, y) becomes | What stays the same |
|---|---|---|
| y-axis | (−x, y) | y-coordinate |
| x-axis | (x, −y) | x-coordinate |
| origin (both axes in turn) | (−x, −y) | distance from O |
From the end-of-chapter exercises: when M is the midpoint of ST, each coordinate of M is the average of the matching coordinates of S and T.
All points at distance r from O lie on a circle with centre O and radius r. Compare the distance d = √x² + y² with r.
Example: A (1, −8), B (−4, 7), C (−7, −4) are all at distance √65 from O, so they lie on circle K of radius √65. D (−5, 6): OD² = 61 < 65, so inside. E (0, 9): OE² = 81 > 65, so outside.
| Idea | Rule |
|---|---|
| Point on x-axis / y-axis | (x, 0) / (0, y) |
| Origin | (0, 0) |
| Quadrant signs | I (+,+), II (−,+), III (−,−), IV (+,−) |
| Distance | √((x₂−x₁)² + (y₂−y₁)²) |
| Reflection in y-axis / x-axis | (−x, y) / (x, −y) |
| Midpoint | ((x₁+x₂)/2, (y₁+y₂)/2) |
Try each problem first, then open the solution to check your working.
State the quadrant, or the axis, of (−4, 7), (5, −2), (−3, −9), (6, 8) and (0, −5).
Look at the signs. (−4, 7): (−, +) so Quadrant II. (5, −2): (+, −) so IV. (−3, −9): (−, −) so III. (6, 8): (+, +) so I. (0, −5): x = 0, so on the y-axis.
Answer: II, IV, III, I, and on the negative y-axis
Find the distance between P (−2, 3) and Q (4, −5).
x₂ − x₁ = 4 − (−2) = 6 and y₂ − y₁ = −5 − 3 = −8.
PQ = √6² + (−8)² = √36 + 64 = √100 = 10.
Answer: 10 units
Find the distances of (−9, 12) and (5, −12) from the origin.
Distance of (x, y) from O is √x² + y².
(−9, 12): √81 + 144 = √225 = 15. (5, −12): √25 + 144 = √169 = 13.
Answer: 15 and 13 units
For what value(s) of x is the distance between (x, 2) and (3, −2) equal to 5?
(x − 3)² + (2 + 2)² = 25, so (x − 3)² = 9 and x − 3 = ±3.
Answer: x = 6 or x = 0
Find the point on the x-axis equidistant from A (2, −5) and B (−2, 9).
Let the point be (x, 0). Then (x − 2)² + 25 = (x + 2)² + 81.
x² − 4x + 29 = x² + 4x + 85, so −8x = 56 and x = −7.
Check: distance to A = √81 + 25 = √106 and to B = √25 + 81 = √106.
Answer: (−7, 0)
A drone is at (−3, 4) on a map grid (1 unit = 1 km) and flies straight to (9, −1). How far does it fly?
Horizontal 12, vertical 5. Distance = √144 + 25 = 13.
Answer: 13 km
In Reiaan’s room the wardrobe has corners W₁ (3, 0) and W₃ (7, 2). How long is its diagonal?
W₁W₃ = √4² + 2² = √20 = 2√5 ≈ 4.47.
Answer: 2√5 ft ≈ 4.47 ft
Show that A (0, 0), B (6, 0), C (3, 4) form an isosceles triangle.
AB = 6. BC = √(3−6)² + 4² = √25 = 5. CA = √3² + 4² = 5.
Two sides are equal (BC = CA = 5). Also 5² + 5² = 50 ≠ 36, so it is not right-angled.
Answer: Isosceles, with BC = CA = 5
Show that P (1, 4), Q (4, 0), R (8, 3) are the vertices of a right isosceles triangle.
PQ = √3² + 4² = 5. QR = √4² + 3² = 5. PR = √7² + 1² = √50.
PQ = QR and PQ² + QR² = 25 + 25 = 50 = PR². By the converse of the Baudhāyana–Pythagoras theorem the angle at Q is 90°.
Answer: Right isosceles, right angle at Q
Are A (−2, −1), B (1, 2), C (4, 5) collinear?
AB = √9 + 9 = 3√2. BC = √9 + 9 = 3√2. AC = √36 + 36 = 6√2.
AB + BC = 6√2 = AC.
Answer: Yes, they lie on one line
Show that (3, 0), (0, 3), (−3, 0), (0, −3) form a square and find its area.
Each side = √3² + 3² = √18. Both diagonals = 6 (along the axes). Four equal sides and equal diagonals give a square.
Area = (√18)² = 18.
Answer: a square of area 18 square units
A (1, 2), B (5, 3), C (6, 6) are three vertices of parallelogram ABCD. Find D.
The diagonals bisect each other. Midpoint of AC = (3.5, 4). This is also the midpoint of BD, so D = (2 × 3.5 − 5, 2 × 4 − 3) = (2, 5).
Check: AB goes (+4, +1) and DC goes (6 − 2, 6 − 5) = (+4, +1).
Answer: D = (2, 5)
Write the reflections of P (3, −4) in the y-axis, the x-axis and the origin. How far is P from each?
y-axis: (−3, −4), distance 2 × 3 = 6. x-axis: (3, 4), distance 2 × 4 = 8. Origin: (−3, 4), distance √6² + 8² = 10.
Answer: (−3, −4), (3, 4), (−3, 4); distances 6, 8, 10
Find the midpoint of (−6, 3) and (4, −9).
x = (−6 + 4)/2 = −1 and y = (3 − 9)/2 = −3.
Answer: (−1, −3)
M (2, −1) is the midpoint of A (−4, 5) and B (x, y). Find B.
(−4 + x)/2 = 2 gives x = 8. (5 + y)/2 = −1 gives y = −7.
Answer: B = (8, −7)
Find the points that divide the segment from A (−3, 6) to B (9, −3) into three equal parts.
The steps from A to B are +12 in x and −9 in y. One third is (+4, −3).
P = (−3 + 4, 6 − 3) = (1, 3) and Q = (1 + 4, 3 − 3) = (5, 0).
Answer: (1, 3) and (5, 0)
A circle has centre O and radius 13. Where do (5, 12), (−12, −5), (9, 9) and (10, 10) lie?
(5, 12): 25 + 144 = 169 = 13², on. (−12, −5): 144 + 25 = 169, on. (9, 9): 162 < 169, inside. (10, 10): 200 > 169, outside.
Answer: on, on, inside, outside
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D₁ is at (8, 0), so the door starts 8 ft from the left wall (the y-axis).
The door lies on the x-axis, so its distance from the x-axis is 0.
D₁ = (8, 0).
Width = 11.5 − 8 = 3.5 ft (about 1.07 m).
A standard room door is about 3 ft (roughly 0.9 m) wide and a wheelchair needs about 2.7 ft (80 cm) clear. So 3.5 ft is comfortable and a wheelchair can enter easily.
Width = 4 − 1.5 = 2.5 ft.
The bathroom door (2.5 ft) is narrower than the room door (3.5 ft).
Room doors are usually about 2.5 to 3 ft (75 to 90 cm) wide. Measure the doors at home and school and compare; main doors are often wider.
Answers will vary. A door is suitable if it is at least about 80 cm clear width and has no step (or has a ramp). Check this for your school’s doors.
0.
Yes. The y-coordinate of every point on the x-axis is 0, so such points are (x, 0).
Yes, but only when x = y. Then Q (y, x) = (x, x) = P. For example P (3, 3) and Q (3, 3). If x ≠ y the points are different.
Yes. Two points are equal only if both coordinates match: x = y and y = x. This holds exactly when x = y. If x ≠ y the first coordinates differ, so the points are different.
Along the x-axis: 7 − 3 = 4 units (to the right). Along the y-axis: 4 − 1 = 3 units (downwards).
Yes. They are the legs of the right triangle ACD with C (3, 1). By the Baudhāyana–Pythagoras theorem AD = √4² + 3² = 5 units.
Same: the y-coordinates, the lengths of all sides (5, √29, √40), the angles and the shape and size.
Changed: the x-coordinates (each changes sign), so the position, and the triangle now faces the other way.
Yes. A reflection in the x-axis sends (x, y) to (x, −y). The x-coordinates stay the same, the y-coordinates change sign, and all the side lengths are preserved again.
A rectangle has the fourth corner with x from the first point and y from the last: (8, 7).
Yes. The table occupies x from 8 to 11 and y from 7 to 9, in the free corner near the right wall. It is clear of the bed (which ends near x = 6), the wardrobe (below y = 2) and the door D₁R₁ on the x-axis.
Length = 11 − 8 = 3 ft; width = 9 − 7 = 2 ft.
The height cannot be found. The plan is a map of the floor, so it shows only length and width.
The door is 2.5 ft wide. Hinged at B₁ (0, 1.5), its free end moves on a circle of radius 2.5 ft about B₁. Its farthest reach into the room is x = 2.5.
The wardrobe starts at x = 3 (corners W₁ (3, 0), W₄ (3, 2)), so the door does not hit it, with 0.5 ft to spare.
If the door is wider than 3 ft it would hit the wardrobe, since B₁ is 3 ft from the wardrobe edge. Suggestions: move the hinge to B₂ (0, 4), which is √3² + 2² ≈ 3.6 ft from the nearest wardrobe corner W₄, or shift the wardrobe to the right, or use a sliding door.
O = (0, 0), F = (0, 9), R = (−6, 9), P = (−6, 0). The bathroom is a rectangle 6 ft wide and 9 ft long.
S = (−6, 6), H = (−3, 6), W = (−2, 9), R = (−6, 9).
SR is vertical (length 3) and SH is horizontal (length 3). SH ∥ RW (both horizontal, RW = 4). HW slants (length √1² + 3² = √10).
So SHWR is a trapezium (a right-angled trapezium) with parallel sides SH = 3 and RW = 4.
Answers will vary. One possible plan:
Washbasin (3 ft along the x-axis, 2 ft up) in the bottom-left corner: (−6, 0), (−3, 0), (−3, 2), (−6, 2).
Toilet (2 ft along x, 3 ft along y) against the left wall: (−6, 3), (−4, 3), (−4, 6), (−6, 6).
PA = 12 − (−6) = 18 ft along the x-axis. The dining room is below the x-axis.
Corners: P (−6, 0), A (12, 0), (12, −15), (−6, −15).
Centre of the dining room = ((−6 + 12)/2, (0 − 15)/2) = (3, −7.5).
Put the 5 ft side along x and the 3 ft side along y. Then x runs from 3 − 2.5 = 0.5 to 5.5 and y from −7.5 − 1.5 = −9 to −6.
Feet: (0.5, −6), (5.5, −6), (5.5, −9), (0.5, −9).
The point is the origin (0, 0). Both coordinates are 0.
A line parallel to the y-axis through W keeps x = −5. So H = (−5, y) for some y; its y-coordinate cannot be predicted.
If y > 0, H is in Quadrant II; if y < 0, H is in Quadrant III. (If y = 0, H is on the x-axis, in no quadrant.)
(i) AM is horizontal (y = −2) and MP is vertical (x = −5), so AM ⊥ MP.
(ii) AM is parallel to the x-axis (MP is parallel to the y-axis).
(iii) M (−5, −2) and P (−5, 2) have equal x and opposite y, so they are mirror images in the x-axis.
Plotting confirms this. (The other sides are RA = √13 and PR = √68.)
Answers may differ. Take I (5, 0) and N (0, −6). Then IZ is vertical and ZN is horizontal, so the right angle is at Z.
IZ = 6, ZN = 5, IN = √5² + 6² = √61 ≈ 7.81.
No. Without negative numbers we could name only points with x ≥ 0 and y ≥ 0, that is Quadrant I and the positive axes. Points to the left of the y-axis or below the x-axis could not be located, so three quarters of the plane would be missing.
Method: find the three distances. Points are collinear if the longest equals the sum of the other two.
MA = √9 + 16 = 5. AG = √36 + 64 = 10. MG = √81 + 144 = 15.
Since MA + AG = 5 + 10 = 15 = MG, the points are collinear.
RB = √9 + 16 = 5. BC = √36 + 49 = √85 ≈ 9.22. RC = √81 + 121 = √202 ≈ 14.21.
Exactly, (5 + √85)² = 110 + 10√85 ≈ 202.2, which is not 202. So RB + BC ≠ RC.
The points are not collinear. (On a graph they look very close to a line, but B is slightly off it.)
Answers may differ. Examples:
(i) O (0, 0), A (4, 0), B (0, 4). The angle at O is 90° and OA = OB = 4.
(ii) O (0, 0), P (−3, −4) in Quadrant III and Q (3, −4) in Quadrant IV. OP = OQ = 5, PQ = 6.
Row 1: midpoint of S, T = ((−3 + 3)/2, 0) = (0, 0) = M. Yes.
Row 2: ((2 + 4)/2, (3 + 5)/2) = (3, 4) = M. Yes.
Row 3: ((0 + 0)/2, (0 − 10)/2) = (0, −5) ≠ M (0, 5). No. (M is on the other side of S; SM = 5, MT = 15.)
Row 4: ((−8 + 6)/2, (7 − 3)/2) = (−1, 2) ≠ M (0, −2). No.
Connection: when M is the midpoint, its x-coordinate is the average of the x-coordinates of S and T, and its y-coordinate is the average of their y-coordinates.
(3 + x)/2 = −7 gives x = −17. (−4 + y)/2 = 1 gives y = 6. So B = (−17, 6).
P is the midpoint of AQ and Q is the midpoint of PB. Let P = (p, q) and Q = (r, s).
x: p = (4 + r)/2 and r = (p + 16)/2. So 4p = 8 + p + 16, giving p = 8 and r = 12.
y: q = (7 + s)/2 and s = (q − 2)/2. So 4q = 14 + q − 2, giving q = 4 and s = 1.
P = (8, 4), Q = (12, 1). Check: each step from A is (+4, −3) and AB = (12, −9) is three such steps.
OA = √1 + 64 = √65. OB = √16 + 49 = √65. OC = √49 + 16 = √65.
All three are at the same distance from O, so they lie on a circle with centre O. Radius = √65 ≈ 8.06.
OD² = 25 + 36 = 61 < 65, so OD < √65: D is inside circle K.
OE² = 0 + 81 = 81 > 65, so OE = 9 > √65: E is outside circle K.
Take D, E, F as the midpoints of BC, CA, AB. Then E + F − D = ((C + A) + (A + B) − (B + C))/2 = A.
A = (6 + 0 − 5, 5 + 3 − 1) = (1, 7). B = D + F − E = (5 + 0 − 6, 1 + 3 − 5) = (−1, −1). C = D + E − F = (5 + 6 − 0, 1 + 5 − 3) = (11, 3).
Check: midpoint of BC = (5, 1) = D; of CA = (6, 5) = E; of AB = (0, 3) = F. ✓
So the vertices are (1, 7), (−1, −1), (11, 3) (the same three points whichever midpoint is assigned to which side).
Draw 10 vertical lines (N–S streets) and 10 horizontal lines (E–W streets), each 1 cm from the next. This gives a square grid 9 cm × 9 cm (1,800 m by 1,800 m) with 100 intersections.
(a) (4, 3) means the 4th N–S street meets the 3rd E–W street: exactly one intersection.
(b) (3, 4) means the 3rd N–S street meets the 4th E–W street: exactly one intersection.
They are different intersections because (4, 3) ≠ (3, 4); the order matters.
(i) Circle A: x from 100 − 80 = 20 to 180; y from 150 − 80 = 70 to 230. Circle B: x from 150 to 350; y from 130 to 330. All values lie in 0 to 800 (x) and 0 to 600 (y), so neither circle goes outside the screen.
(ii) AB = √150² + 80² = √28900 = 170. The radii add to 80 + 100 = 180 and differ by 20. Since 20 < 170 < 180, the circles intersect (at two points).
AB = √9 + 1 = √10. BC = √1 + 9 = √10. CD = √9 + 1 = √10. DA = √1 + 9 = √10. All four sides are equal.
Diagonals: AC = √16 + 4 = √20 and BD = √4 + 16 = √20. They are equal. A four-sided figure with equal sides and equal diagonals is a square.
(Check the angle: AB² + BC² = 20 = AC², so the angle at B is 90°.)
Area = (√10)² = 10 square units.