NCERT Exercise Questions & Answers + Important Extra Questions
Exercise 7.1
Q1. Find the distance between the following pairs of points
Distance formula: d = √[(x₂–x₁)² + (y₂–y₁)²]
(i) (2, 3) and (4, 1): d = √[(4–2)² + (1–3)²] = √[4 + 4] = 2√2
(ii) (–5, 7) and (–1, 3): d = √[(–1+5)² + (3–7)²] = √[16 + 16] = 4√2
(iii) (a, b) and (–a, –b): d = √[(–a–a)² + (–b–b)²] = √[4a² + 4b²] = 2√(a² + b²)
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B?
Answer
d = √(36² + 15²) = √(1296 + 225) = √1521 = 39 units. Yes, the distance between towns A and B is 39 km.
Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear.
Answer
AB = √[(2–1)² + (3–5)²] = √5. BC = √[(–2–2)² + (–11–3)²] = √(16+196) = √212. AC = √[(–2–1)² + (–11–5)²] = √(9+256) = √265.
AB + BC = √5 + √212 ≠ √265 = AC. Not collinear.
Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.
Answer
AB = √(1 + 36) = √37. BC = √(1 + 36) = √37. AC = √(4 + 0) = 2.
AB = BC = √37 ≠ AC. Yes, isosceles triangle (two sides equal).
Name the type of quadrilateral formed by the vertices (–1, –2), (1, 0), (–1, 2), (–3, 0).
Answer
AB = √(4+4) = 2√2, BC = √(4+4) = 2√2, CD = √(4+4) = 2√2, DA = √(4+4) = 2√2.
All sides equal. Diagonals: AC = √(0+16) = 4, BD = √(16+0) = 4. Diagonals equal.
All sides equal + diagonals equal
⇒ Square
Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
Answer
Point on x-axis = (x, 0). Distance from (2,–5) = Distance from (–2,9)
(x–2)² + 25 = (x+2)² + 81
⇒ x² – 4x + 4 + 25 = x² + 4x + 4 + 81
⇒ –8x = 56
⇒ x = –7
Point = (–7, 0)
Find the values of y for which the distance between (2, –3) and (10, y) is 10 units.
Answer
√[(10–2)² + (y+3)²] = 10
⇒ 64 + (y+3)² = 100
⇒ (y+3)² = 36
⇒ y + 3 = ±6
y = 3 or y = –9
Find a relation between x and y such that (x, y) is equidistant from (3, 6) and (–3, 4).
Answer
(x–3)² + (y–6)² = (x+3)² + (y–4)²
x² – 6x + 9 + y² – 12y + 36 = x² + 6x + 9 + y² – 8y + 16
–12x – 4y + 20 = 0 → 3x + y – 5 = 0
Exercise 7.2
Find the coordinates of the point which divides the line segment joining (4, –3) and (8, 5) in the ratio 3:1 internally.
Answer
Section formula: x = (m×x₂ + n×x₁)/(m+n), y = (m×y₂ + n×y₁)/(m+n)
x = (3×8 + 1×4)/4 = 28/4 = 7, y = (3×5 + 1×(–3))/4 = 12/4 = 3
Point = (7, 3)
Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
Answer
Two points divide in ratios 1:2 and 2:1.
P₁ (1:2): x = (1×(–2) + 2×4)/3 = 6/3 = 2, y = (1×(–3) + 2×(–1))/3 = –5/3. P₁ = (2, –5/3)
P₂ (2:1): x = (2×(–2) + 1×4)/3 = 0/3 = 0, y = (2×(–3) + 1×(–1))/3 = –7/3. P₂ = (0, –7/3)
Find the ratio in which the line segment joining (–3, 10) and (6, –8) is divided by (–1, 6).
Answer
Let ratio = k:1. –1 = (6k – 3)/(k+1)
⇒ –k – 1 = 6k – 3
⇒ 2 = 7k
⇒ k = 2/7.
Ratio = 2:7
Find the ratio in which the line segment joining A(1, –5) and B(–4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Answer
On x-axis, y = 0. Let ratio = k:1. y = (5k + (–5))/(k+1) = 0
⇒ 5k = 5
⇒ k = 1.
Ratio = 1:1 (midpoint). x = (–4 + 1)/2 = –3/2. Point = (–3/2, 0)
If A(–2, –1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram, find a and b.
Answer
Diagonals of a parallelogram bisect each other. Midpoint of AC = Midpoint of BD.
((–2+4)/2, (–1+b)/2) = ((a+1)/2, (0+2)/2)
1 = (a+1)/2
⇒ a = 1. (–1+b)/2 = 1
⇒ b = 3. a = 1, b = 3
Exercise 7.3
Q1. Find the area of the triangle whose vertices are
Area = ½|x₁(y₂–y₃) + x₂(y₃–y₁) + x₃(y₁–y₂)|
(i) (2, 3), (–1, 0), (2, –4): A = ½|2(0+4) + (–1)(–4–3) + 2(3–0)| = ½|8 + 7 + 6| = 21/2 = 10.5 sq units
(ii) (–5, –1), (3, –5), (5, 2): A = ½|–5(–5–2) + 3(2+1) + 5(–1+5)| = ½|35 + 9 + 20| = 32 sq units
In each of the following find the value of 'k', for which the points are collinear:
(i) (7, –2), (5, 1), (3, k): Area = 0 → 7(1–k) + 5(k+2) + 3(–2–1) = 0 → 7–7k+5k+10–9 = 0 → –2k + 8 = 0 → k = 4
(ii) (8, 1), (k, –4), (2, –5): 8(–4+5) + k(–5–1) + 2(1+4) = 0 → 8 – 6k + 10 = 0 → k = 3
Answer
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Answer
Mid-points: M₁ = (1, 0), M₂ = (1, 2), M₃ = (0, 1).
Area of mid-point triangle = ½|1(2–1) + 1(1–0) + 0(0–2)| = ½|1+1| = 1 sq unit
Area of original = ½|0(1–3) + 2(3+1) + 0(–1–1)| = ½|8| = 4 sq units.
Ratio = 1:4. The area of the triangle formed by mid-points is 1/4 of the original.
Find the area of the quadrilateral whose vertices taken in order are (–4, –2), (–3, –5), (3, –2), (2, 3).
Answer
Divide into two triangles by diagonal from (–4,–2) to (3,–2).
Area of Δ₁ with (–4,–2), (–3,–5), (3,–2) = ½|–4(–5+2) + (–3)(–2+2) + 3(–2+5)| = ½|12 + 0 + 9| = 21/2
Area of Δ₂ with (–4,–2), (3,–2), (2,3) = ½|–4(–2–3) + 3(3+2) + 2(–2+2)| = ½|20 + 15 + 0| = 35/2
Total = 21/2 + 35/2 = 28 sq units
Important Extra Questions
Find the centroid of the triangle with vertices (3, –5), (–7, 4) and (10, –2).
Answer
Centroid G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3) = ((3–7+10)/3, (–5+4–2)/3) = (2, –1)
If the centroid of the triangle formed by (7, –6), (2, –2) and (h, k) is (4, –3), find h and k.
Answer
(7+2+h)/3 = 4
⇒ h = 3. (–6–2+k)/3 = –3
⇒ k = –1. h = 3, k = –1
Show that the points A(1, 2), B(5, 4), C(3, 8) and D(–1, 6) are vertices of a square.
Answer
AB = √(16+4) = √20. BC = √(4+16) = √20. CD = √(16+4) = √20. DA = √(4+16) = √20.
All sides equal. Diagonal AC = √(4+36) = √40. BD = √(36+4) = √40. Diagonals equal.
All sides equal + diagonals equal
⇒ ABCD is a square.