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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