These NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-4-ex-4-3/

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

**Ex 4.3 Class 12 NCERT Solutions 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:

**Exercise 4.3 Class 12 NCERT Solutions Question 2.**

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

Solution:

∴ The given points are collinear.

**4.3 Class 12 NCERT Solutions 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)

\(\frac{1}{2}\left|\begin{array}{ccc} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{array}\right|\) =

= ± 4 ⇒ \(\frac{1}{2}\)[- 2(4 – k) – 0 + (0)] = ± 4

⇒ 8 + 2k = ± 8 ⇒ 2k = 16 or 2k = 0 or k = 8

**Exercise 4.3 Class 12 Maths NCERT Solutions 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) Let (x, y) be a point on the line, then the points (x, y), (1, 2) and (3, 6) are collinear.

∴ \(\left|\begin{array}{lll}

x & y & 1 \\

1 & 2 & 1 \\

3 & 6 & 1

\end{array}\right|\) = 0

⇒ x(2 – 6) – y(1 – 3) + 1(6 – 6) = 0

⇒ – 4x + 2y = 0 ⇒ 2x – y = 0

⇒ y = 2x

(ii) Let (x, y) be a point on the line. Then (r, y), (3, 1) and (9, 3) are collinear.

∴ \(\left|\begin{array}{lll}

x & y & 1 \\

3 & 1 & 1 \\

9 & 3 & 1

\end{array}\right|\) = 0

⇒ x(1 – 3) – y(3 – 9) + 1(9 – 9) = 0

⇒ 2x + 6y = 0 ⇒ x – 3y = 0

**Ex 4.3 Class 12 Maths Ncert Solutions 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