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NCERT Solutions For Class 12 Maths Chapter 4 Determinants Ex 4.3

Here, Below you all know about NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3 Question Answer. I know many of you confuse about finding Chapter 4 Determinants Ex 4.3 Of Class 12 NCERT Solutions. So, Read the full post below and get your solutions.

TextbookNCERT
BoardCBSE
CategoryNCERT Solutions
ClassClass 12
SubjectMaths
ChapterChapter 4
ExerciseClass 12 Maths Chapter 4 Determinants Exercise 4.3
Number of Questions Solved5
NCERT Solutions For Class 12 Maths Chapter 4 Determinants Ex 4.3

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3

NCERT TEXTBOOK EXERCISES

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 = $\frac { 1 }{ 2 } \left| \begin{matrix} 1\quad & 0 & \quad 1 \ 6\quad & 0 & \quad 1 \ 4\quad & 3 & \quad 1 \end{matrix} \right|$
= 12 [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)
$\frac { 1 }{ 2 } \left| \begin{matrix} k\quad & 0 & \quad 1 \ 4\quad & 0 & \quad 1 \ 0\quad & 2 & \quad 1 \end{matrix} \right|$
= 12 [-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 ∆ = $\frac { 1 }{ 2 } \left| \begin{matrix} 2\quad & -6 & \quad 1 \ 5\quad & 4 & \quad 1 \ k\quad & 4 & \quad 1 \end{matrix} \right|$
= 12 [50 – 10k] = 25 – 5k
∴ 25-5k = 35 or 25-5k = -35
-5k = 10 or 5k = 60
=> k = -2 or k = 12

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

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