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# 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 R_{3} 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 (x_{1}, y_{1}), (x_{2}, y_{2}), and (x_{3}, y_{3}) is the absolute value of the determinant (âˆ†), 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.