# NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3

In this chapter, we provide NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3 for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3 pdf, free NCERT solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3 book pdf download.

## 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 = $frac { 1 }{ 2 } left| begin{matrix} 1quad & 0 & quad 1 \ 6quad & 0 & quad 1 \ 4quad & 3 & quad 1 end{matrix} right|$
= $\ frac { 1 }{ 2 }$ [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) $frac { 1 }{ 2 } left| begin{matrix} kquad & 0 & quad 1 \ 4quad & 0 & quad 1 \ 0quad & 2 & quad 1 end{matrix} right|$
= $\ frac { 1 }{ 2 }$ [-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 ∆ = $frac { 1 }{ 2 } left| begin{matrix} 2quad & -6 & quad 1 \ 5quad & 4 & quad 1 \ kquad & 4 & quad 1 end{matrix} right|$
= $\ frac { 1 }{ 2 }$ [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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