NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths 4 Determinants Ex 4.3.
- Determinants Class 12 Ex 4.1
- Determinants Class 12 Ex 4.2
- Determinants Class 12 Ex 4.4
- Determinants Class 12 Ex 4.5
- Determinants Class 12 Ex 4.6
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 12 |
| Subject | Maths |
| Chapter | Chapter 4 |
| Chapter Name | Determinants |
| Exercise | Ex 4.3 |
| Number of Questions Solved | 5 |
| Category | NCERT Solutions |
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3
Ex 4.3 Class 12 Maths Question 1.
Find the 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) Area of triangle = \(\frac { 1 }{ 2 } \left| \begin{matrix} 1\quad & 0 & \quad 1 \\ 6\quad & 0 & \quad 1 \\ 4\quad & 3 & \quad 1 \end{matrix} \right| \)
= \(\\ \frac { 1 }{ 2 } \) [1(0-3)+1(18-0)]
= 7.5 sq units
Ex 4.3 Class 12 Maths Question 2.
Show that the points A (a, b + c), B (b, c + a) C (c, a+b) are collinear.
Solution:
The vertices of ∆ABC are A (a, b + c), B (b, c + a) and C (c, a + b)
Ex 4.3 Class 12 Maths Question 3.
Find the value 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:
(i) Area of ∆ = 4 (Given)
\(\frac { 1 }{ 2 } \left| \begin{matrix} k\quad & 0 & \quad 1 \\ 4\quad & 0 & \quad 1 \\ 0\quad & 2 & \quad 1 \end{matrix} \right| \)
= \(\\ \frac { 1 }{ 2 } \) [-2k+8]
= -k+4
Case (a): -k + 4 = 4 ==> k = 0
Case(b): -k + 4 = -4 ==> k = 8
Hence, k = 0,8
(ii) The area of the triangle whose vertices are (-2,0), (0,4), (0, k)
Ex 4.3 Class 12 Maths Question 4.
(i) Find the equation of line joining (1, 2) and (3,6) using determinants.
(ii) Find the equation of line joining (3,1), (9,3) using determinants.
Solution:
(i) Given: Points (1,2), (3,6)
Equation of the line is
Ex 4.3 Class 12 Maths Question 5.
If area of triangle is 35 sq. units with vertices (2, – 6), (5,4) and (k, 4). Then k is
(a) 12
(b) – 2
(c) -12,-2
(d) 12,-2
Solution:
(d) Area of ∆ = \(\frac { 1 }{ 2 } \left| \begin{matrix} 2\quad & -6 & \quad 1 \\ 5\quad & 4 & \quad 1 \\ k\quad & 4 & \quad 1 \end{matrix} \right| \)
= \(\\ \frac { 1 }{ 2 } \) [50 – 10k] = 25 – 5k
∴ 25-5k = 35 or 25-5k = -35
-5k = 10 or 5k = 60
=> k = -2 or k = 12
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