NCERT Solution Exercise 3
Question 1: Find area of the triangle with vertices at the point given in each of the following:
(i) (1, 0), (6, 0), (4, 3) (ii) (2, 7), (1, 1), (10, 8)
(iii) (−2, −3), (3, 2), (−1, −8)
Answer
(i) The area of the triangle with vertices (1, 0), (6, 0), (4, 3) is given by the relation,
= ½ [ 1(0 - 3) – 0(6 - 4) + 1(18 – 0)]
= ½ [ -3 + 18] = 15/2 square units
(ii) The area of the triangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,
= ½ [ 2(1 - 8) – 7(1 - 10) + 1(8 – 10)]
= ½ [ 2(- 7) – 7(- 9) + 1(– 2)]
= ½ [ -14 + 63 – 2] = ½ [-16 + 63]
= 47/2 square units
(iii)The area of the triangle with vertices (−2, −3), (3, 2), (−1, −8) is given by the relation,
= ½ [ 2(1 - 8) – 7(1 - 10) + 1(8 – 10)]
= ½ [ -2(10) + 3(4) + 1(– 22)]
= ½ [ -20 + 12 – 22]
= - (30/2) = -15
Hence, the area of the triangle is |-15|= 15 square units.
Question 2: Show that points A (a, b + c), B(b, c + a), C(c, a+b)
are collinear
Answer
Area of ΔABC is given by the relation,
= 0 (All electric of R3 are 0)
Thus, the area of the triangle formed by points A, B, and C is zero.
Hence, the points A, B, and C are collinear.
Question 3: Find values of k if area of triangle is 4 square units and vertices are
(i) (k, 0), (4, 0), (0, 2) (ii) (−2, 0), (0, 4), (0, k)
Answer
We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is the absolute value of the determinant (∆), where
It is given that the area of triangle is 4 square units.
∴ Δ = ± 4.
(i) The area of the triangle with vertices (k, 0), (4, 0), (0, 2) is given by the relation,
= ½ [k (0 - 2) – 0 (4 - 0) + 1(8 - 0)]
= ½ [-2k + 8] = -k + 4
∴ −k + 4 = ± 4
When −k + 4 = − 4, k = 8.
When −k + 4 = 4, k = 0.
Hence, k = 0, 8.
(ii) The area of the triangle with vertices (−2, 0), (0, 4), (0, k) is given by the relation,
= ½ [-2(4 - k)]
= k – 4
∴ k − 4 = ± 4
When k − 4 = − 4, k = 0.
When k − 4 = 4, k = 8.
Hence, k = 0, 8.
Question 4:
(i) Find equation of line joining (1, 2) and (3, 6) using determinants
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants
Answer
(i) Let P (x, y) be any point on the line joining points A (1, 2) and B (3, 6). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.
⇒ ½ [1 (6 - y) – 2 (3 - x) + 1(3y – 6x)] = 0
⇒ 6 – y – 6 + 2x + 3y – 6x = 0
⇒ 2y – 4x = 0
⇒ y = 2x
Hence, the equation of the line joining the given points is y = 2x.
(ii) Let P (x, y) be any point on the line joining points A (3, 1) and B (9, 3). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.
⇒ ½ [3 (3 - y) – 1 (9 - x) + 1(9y – 3x)] = 0
⇒ 9 – 3y – 9 + x + 9y – 3x = 0
⇒ 6y – 2x = 0
⇒ x – 3y = 0
Hence, the equation of the line joining the given points is x − 3y = 0.
Question 5: If area of triangle is 35 square units with vertices (2, −6), (5, 4), and (k, 4). Then k is
A. 12 B. −2 C. −12, −2 D. 12, −2
Answer: D
The area of the triangle with vertices (2, −6), (5, 4), and (k, 4) is given by the relation,
= ½ [2 (4 - 4) + 6 (5 - k) + 1(20 – 4k)]
= ½ [30 – 6k + 20 – 4k]
= ½ [50 – 10k]
= 25 – 5k
It is given that the area of the triangle is ±35.
Therefore, we have:
⇒ 25 – 5k = ±35
⇒ 5(5 - k) = ±35
⇒ 5 - k = ±7
When 5 − k = −7, k = 5 + 7 = 12.
When 5 − k = 7, k = 5 − 7 = −2.
Hence, k = 12, −2.
The correct answer is D.