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.
Using coordinates, we can find the distance between two points and the point that divides a segment in a given ratio. These let us test shapes (isosceles, square, parallelogram), check collinearity, and locate points.
The distance between P(x₁, y₁) and Q(x₂, y₂) is
Distance from the origin: OP = √x² + y².
(x₂ − x₁)² = (x₁ − x₂)², so the order of the points does not matter.
| To show | Check |
|---|---|
| Three points are collinear | the sum of two distances equals the third |
| Isosceles triangle | two sides equal |
| Equilateral triangle | all three sides equal |
| Right triangle | square of the longest side = sum of squares of the other two |
| Square | four equal sides and equal diagonals |
| Rhombus | four equal sides, diagonals unequal |
| Rectangle | opposite sides equal and equal diagonals |
| Parallelogram | opposite sides equal (or diagonals bisect each other) |
If P is equidistant from A and B, write PA² = PB². Squaring removes the roots. The point on the x-axis equidistant from A and B is (x, 0); solve for x.
The point P that divides the segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂ is
The first part of the ratio goes with the second point. AP : PB = m₁ : m₂.
The mid-point (ratio 1 : 1) of the join of A(x₁, y₁) and B(x₂, y₂) is
To find an unknown ratio, take it as k : 1. Then P = ((kx₂ + x₁)/(k + 1), (ky₂ + y₁)/(k + 1)). Equate one coordinate to find k, and check with the other.
| Quantity | Formula |
|---|---|
| Distance PQ | √[(x₂ − x₁)² + (y₂ − y₁)²] |
| Distance from origin | √(x² + y²) |
| Section (m₁ : m₂) | ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)) |
| Mid-point | ((x₁ + x₂)/2, (y₁ + y₂)/2) |
Try each problem first, then open the solution to check your working.
Find the distance between (−1, 3) and (2, 7).
Use: PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]
√(3² + 4²) = √25
Answer: 5 units
How far is the point (3, −4) from the origin?
Use: √(x² + y²)
√(9 + 16) = 5
Answer: 5 units
Show that (1, 2), (4, 6) and (7, 2) form an isosceles triangle.
Given: A(1, 2), B(4, 6), C(7, 2)
AB = √(9 + 16) = 5; BC = √(9 + 16) = 5; AC = √(36 + 0) = 6
AB = BC
Answer: Isosceles, with AB = BC = 5
Show that A(1, 1), B(4, 4) and C(3, 5) form a right triangle.
Use: Pythagoras
AB² = 9 + 9 = 18; BC² = 1 + 1 = 2; AC² = 4 + 16 = 20
AB² + BC² = 20 = AC², so the right angle is at B.
Answer: Right-angled at B
Are (1, −1), (5, 2) and (9, 5) collinear?
Given: A, B, C
Use: AB + BC = AC?
AB = √(16 + 9) = 5; BC = √(16 + 9) = 5; AC = √(64 + 36) = 10
AB + BC = AC
Answer: Yes, they are collinear
Find x if the distance between (x, 2) and (3, −2) is 5.
Use: Square both sides
(x − 3)² + 16 = 25 ⇒ (x − 3)² = 9 ⇒ x − 3 = ±3
Answer: x = 6 or x = 0
Find the point on the y-axis equidistant from A(5, 2) and B(−4, 3).
Given: P(0, y)
Use: PA² = PB²
25 + (y − 2)² = 16 + (y − 3)² ⇒ 25 − 4y + 4 = 16 − 6y + 9 ⇒ 2y = −4 ⇒ y = −2
Answer: (0, −2)
Show that (2, −2), (8, 4), (5, 7) and (−1, 1) are the vertices of a rectangle.
Given: A, B, C, D
AB = √72, BC = √18, CD = √72, DA = √18: opposite sides equal.
AC = √(9 + 81) = √90, BD = √(81 + 9) = √90: diagonals equal.
Answer: It is a rectangle
Find the point that divides the join of (2, −3) and (8, 9) in the ratio 2 : 1.
Given: m₁ = 2, m₂ = 1
Use: Section formula
x = (2 × 8 + 1 × 2)/3 = 6
y = (2 × 9 + 1 × (−3))/3 = 5
Answer: (6, 5)
Find the mid-point of (−1, 2) and (5, −4).
Use: ((x₁ + x₂)/2, (y₁ + y₂)/2)
(4/2, −2/2)
Answer: (2, −1)
In what ratio does (4, 1) divide the join of (2, 3) and (10, −5)?
Given: ratio k : 1
x: (10k + 2)/(k + 1) = 4 ⇒ 10k + 2 = 4k + 4 ⇒ k = 1/3
Check y: (−5/3 + 3)/(4/3) = (4/3)/(4/3) = 1 ✓
Answer: 1 : 3
In what ratio does the y-axis divide the join of (−4, 0) and (8, 6)? Find the point.
Given: ratio k : 1; on the y-axis x = 0
(8k − 4)/(k + 1) = 0 ⇒ k = 1/2
y = (6 × 1/2 + 0)/(3/2) = 2
Answer: 1 : 2; the point is (0, 2)
The centre of a circle is (2, −1) and one end of a diameter is (−3, 4). Find the other end.
Use: Centre = mid-point of the diameter
(−3 + x)/2 = 2 ⇒ x = 7; (4 + y)/2 = −1 ⇒ y = −6
Answer: (7, −6)
(1, 0), (4, 2) and (5, 6) are three consecutive vertices of a parallelogram. Find the fourth.
Given: A(1, 0), B(4, 2), C(5, 6), D(x, y)
Use: Diagonals bisect each other
Mid-point of AC = (3, 3) = mid-point of BD = ((4 + x)/2, (2 + y)/2)
x = 2, y = 4
Answer: (2, 4)
The medians of a triangle meet at the centroid, which divides each median in the ratio 2 : 1 from the vertex. Find the centroid of (1, 2), (5, −1) and (6, 8).
Use: Mid-point of BC, then 2 : 1
Mid-point of BC: M = (11/2, 7/2)
Centroid = ((2 × 11/2 + 1)/3, (2 × 7/2 + 2)/3) = (12/3, 9/3)
Answer: (4, 3), which is also ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3)
Find the points of trisection of the join of (−3, 6) and (9, −3).
Given: ratios 1 : 2 and 2 : 1
1 : 2 → ((9 − 6)/3, (−3 + 12)/3) = (1, 3)
2 : 1 → ((18 − 3)/3, (−6 + 6)/3) = (5, 0)
Answer: (1, 3) and (5, 0)
Show that (2, 3), (7, 3), (7, −2) and (2, −2) form a square.
Each side has length 5.
Diagonals: (2, 3) to (7, −2) is √50; (7, 3) to (2, −2) is √50.
Answer: Equal sides and equal diagonals: a square
Find the length of the median through A in the triangle A(−1, 3), B(1, −1), C(5, 1).
Use: Mid-point of BC, then distance
Mid-point of BC = (3, 0)
AM = √(16 + 9) = 5
Answer: 5 units
Two vertices of an equilateral triangle are (2, 0) and (−2, 0). Find the third vertex on the positive y-axis.
Given: third vertex (0, y)
Use: All sides equal
Side = 4. Distance from (0, y) to (2, 0): 4 + y² = 16 ⇒ y = 2√3 (positive).
Answer: (0, 2√3)
Find the area of the rhombus with vertices (2, 3), (5, 7), (8, 3) and (5, −1).
Use: Area = ½ × product of diagonals
Diagonal (2, 3)–(8, 3) = 6; diagonal (5, 7)–(5, −1) = 8
Area = ½ × 6 × 8
Answer: 24 square units
Pick a level and answer 10 questions with instant feedback. Each level has a bank of 20 questions, and New 10 questions draws a fresh set.
Read the Assertion (A) and the Reason (R), then pick one of the four standard options. You get 5 questions per set from a bank of 20 at each level, and New 5 questions draws a fresh set.
Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.
(i) √(4 − 2)² + (1 − 3)² = √4 + 4 = 2√2
(ii) √(−1 + 5)² + (3 − 7)² = √16 + 16 = 4√2
(iii) √(−a − a)² + (−b − b)² = √4a² + 4b² = 2√a² + b²
Answer: (i) 2√2 (ii) 4√2 (iii) 2√(a² + b²)
Distance = √36² + 15² = √1296 + 225 = √1521 = 39.
Town A is at the origin and town B at (36, 15), with 1 km as one unit, so the towns are 39 km apart.
Answer: 39 units; the towns are 39 km apart
Let A(1, 5), B(2, 3), C(−2, −11). AB = √1 + 4 = √5, BC = √16 + 196 = √212, AC = √9 + 256 = √265.
√5 ≈ 2.24, √212 ≈ 14.56, √265 ≈ 16.28. No two add up to the third (2.24 + 14.56 = 16.80 ≠ 16.28).
Answer: Not collinear
A(5, −2), B(6, 4), C(7, −2). AB = √1 + 36 = √37, BC = √1 + 36 = √37, AC = √4 + 0 = 2.
AB = BC, and the points are not collinear.
Answer: Yes, it is an isosceles triangle
From the figure, A(3, 4), B(6, 7), C(9, 4), D(6, 1).
AB = √9 + 9 = 3√2, BC = √9 + 9 = 3√2, CD = √9 + 9 = 3√2, DA = √9 + 9 = 3√2.
Diagonals: AC = √36 + 0 = 6, BD = √0 + 36 = 6.
All sides are equal and the diagonals are equal, so ABCD is a square.
Answer: Champa is correct
(i) A(−1, −2), B(1, 0), C(−1, 2), D(−3, 0). AB = BC = CD = DA = √8 = 2√2. Diagonals AC = 4, BD = 4. Equal sides and equal diagonals: a square.
(ii) A(−3, 5), B(3, 1), C(0, 3), D(−1, −4). AC = √9 + 4 = √13, CB = √9 + 4 = √13, AB = √36 + 16 = 2√13. AC + CB = AB, so A, C, B are collinear. No quadrilateral is formed.
(iii) A(4, 5), B(7, 6), C(4, 3), D(1, 2). AB = √10, BC = √18, CD = √10, DA = √18: opposite sides equal. Diagonals AC = 2, BD = √52: unequal. So it is a parallelogram (not a rectangle).
Answer: (i) square (ii) no quadrilateral (iii) parallelogram
Let the point be (x, 0). (x − 2)² + 25 = (x + 2)² + 81.
x² − 4x + 29 = x² + 4x + 85 ⇒ −8x = 56 ⇒ x = −7.
Answer: (−7, 0)
(10 − 2)² + (y + 3)² = 100 ⇒ (y + 3)² = 36 ⇒ y + 3 = ±6.
Answer: y = 3 or y = −9
QP² = 25 + 16 = 41. QR² = x² + 25. So x² + 25 = 41 ⇒ x = ±4.
QR = √41 in both cases.
If x = 4: PR = √(4 − 5)² + (6 + 3)² = √82. If x = −4: PR = √81 + 81 = 9√2.
Answer: x = ±4; QR = √41; PR = √82 (x = 4) or 9√2 (x = −4)
(x − 3)² + (y − 6)² = (x + 3)² + (y − 4)²
−6x + 9 − 12y + 36 = 6x + 9 − 8y + 16 ⇒ −12x − 4y + 20 = 0.
Answer: 3x + y − 5 = 0
x = 2(4) + 3(−1)5 = 1, y = 2(−3) + 3(7)5 = 3.
Answer: (1, 3)
Ratio 1 : 2: (1(−2) + 2(4)3, 1(−3) + 2(−1)3) = (2, −53).
Ratio 2 : 1: (2(−2) + 1(4)3, 2(−3) + 1(−1)3) = (0, −73).
Answer: (2, −5/3) and (0, −7/3)
Take A as the origin, AB along the x-axis and AD along the y-axis. AD = 100 m.
Green flag: on line 2, at ¼ × 100 = 25 m: (2, 25). Red flag: on line 8, at ⅕ × 100 = 20 m: (8, 20).
Distance = √6² + 5² = √61 m.
Blue flag at the mid-point: ((2 + 8)/2, (25 + 20)/2) = (5, 22.5), i.e. on the 5th line, 22.5 m from AD's start.
Answer: √61 m apart; the blue flag goes on the 5th line at 22.5 m
Let the ratio be k : 1. x: 6k − 3k + 1 = −1 ⇒ 6k − 3 = −k − 1 ⇒ k = 27.
Check y: −8(2/7) + 102/7 + 1 = 54/79/7 = 6 ✓
Answer: 2 : 7
Let the ratio be k : 1. On the x-axis, y = 0: 5k − 5k + 1 = 0 ⇒ k = 1.
x = −4 + 12 = −32.
Answer: 1 : 1; the point is (−3/2, 0)
The diagonals of a parallelogram bisect each other, so the mid-points of the diagonals joining (1, 2) to (x, 6) and (4, y) to (3, 5) are the same.
(1 + x2, 4) = (72, y + 52) ⇒ x = 6 and y = 3.
Answer: x = 6, y = 3
The centre is the mid-point of AB. Let A = (x, y): x + 12 = 2 ⇒ x = 3; y + 42 = −3 ⇒ y = −10.
Answer: A(3, −10)
AP = 3/7 AB means PB = 4/7 AB, so AP : PB = 3 : 4.
x = 3(2) + 4(−2)7 = −27, y = 3(−4) + 4(−2)7 = −207.
Answer: P(−2/7, −20/7)
The points divide AB in the ratios 1 : 3, 1 : 1 and 3 : 1.
1 : 3: (2 − 64, 8 + 64) = (−1, 72)
1 : 1: (0, 5)
3 : 1: (6 − 24, 24 + 24) = (1, 132)
Answer: (−1, 7/2), (0, 5), (1, 13/2)
Diagonal from (3, 0) to (−1, 4): √16 + 16 = 4√2.
Diagonal from (4, 5) to (−2, −1): √36 + 36 = 6√2.
Area = 12 × 4√2 × 6√2 = 24.
Answer: 24 square units