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