These NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4.3
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:
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.
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
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
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