**Class 12 Maths NCERT Solutions for Chapter 4 Determinants Exercise 4.3**

**Determinants Exercise 4.3 Solutions**

**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) **

**Solution**

(i) The area of the triangle with vertices (1, 0), (6, 0), (4, 3) is given by the relation,

= (1/2) [1(0 - 3) - 0(6 - 4) + 1(18 - 0)]

= (1/2)[-3 + 18]

= (1/2) [15]

= 15/2

Hence, area of the triangle is 15/2 square units.

(ii) The area of the triangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,

= (1/2)[2(1-8) -7(1-10) +1(8-10)]

= (1/2)[ -14 + 63 - 2]

= (1/2)[47]

= 47/2

Hence, area of the triangle is 47/2 square units.

(iii) The area of the triangle with vertices (-2, -3) , (3, 2), (-1, -8) is given by the relation,

= (1/2)[-2(2 + 8) + 3(3 + 1) + 1(-24 + 2)]

= (1/2)[-2(10) + 3(4) + 1(-22)]

= (1/2)[-20 + 12 - 22]

= (1/2) [-30]

= -30/2 = -15

Hence, area of the triangle is 15 square units.

**2. Show that the points A(a, b + c), B(b, c + a), C(c, a + b) are collinear. **

**Solution**

The area of the triangle with vertices A(a, b + c), B(b, c + a), C(c, a + b) is given by the absolute value of the relation :

Thus, the area of the triangle formed by points A, B, and C is zero.

Hence, the points A, B and C are collinear.

**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) **

**Solution**

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 (Î”), where

If is given that the area of triangle is 4 square units.

∴ Î”＝± 4.

(i) The are of the triangle with vertices (k, 0), (4, 0) , (0, 2) is given by the relation,

= (1/2)[k(0 - 2) - 0(4 - 0) + 1(8 - 0)]

= (1/2) [-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 are of the triangle with vertices (-2, 0), (0, 4), (0, k) is given by the relation,

= (1/2) [-2(4 - k)

= k - 4

∴ k - 4 = ± 4

When k - 4 = - 4 , k = 0

When k - 4 = 4, k = 8

Hence, k = 0, 8

**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. **

**Solution**

(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.

⇒ 1/2 [ 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.

⇒ 1/2 [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.

**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

**Solution**

The area of the triangle with vertices (2, -6), (5, 4), and (k, 4) is given by the relation ,

= 1/2 [2(4 - 4) + 6(5 - k) + 1(20 -4k)]

= 1/2 [30 - 6k + 20 - 4k ]

= 1/2 [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.