NCERT Solutions Class-12-Chapter-4-Determinants
Excercise-4.2
Note: The area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by the absolute value of the determinant: \[ \Delta = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \] Three points are collinear if the area of the triangle formed by them is zero.
Q1
Find 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) \( (-2, -3), (3, 2), (-1, -8) \)
(i) \( (1, 0), (6, 0), (4, 3) \)
(ii) \( (2, 7), (1, 1), (10, 8) \)
(iii) \( (-2, -3), (3, 2), (-1, -8) \)
Solution:
(i) Vertices: \( (1, 0), (6, 0), (4, 3) \)
\[ \Delta = \frac{1}{2} \begin{vmatrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix} \]
Expanding along the second column:
\[ \Delta = \frac{1}{2} [-0 + 0 - 3(1-6)] = \frac{1}{2} [-3(-5)] = \frac{15}{2} \text{ sq units} \]
(ii) Vertices: \( (2, 7), (1, 1), (10, 8) \)
\[ \Delta = \frac{1}{2} \begin{vmatrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{vmatrix} \]
Expanding along R1:
\[ \begin{aligned} \Delta &= \frac{1}{2} [2(1-8) - 7(1-10) + 1(8-10)] \\ &= \frac{1}{2} [2(-7) - 7(-9) + 1(-2)] \\ &= \frac{1}{2} [-14 + 63 - 2] = \frac{47}{2} \text{ sq units} \end{aligned} \]
(iii) Vertices: \( (-2, -3), (3, 2), (-1, -8) \)
\[ \Delta = \frac{1}{2} \begin{vmatrix} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \end{vmatrix} \]
\[ \begin{aligned} \Delta &= \frac{1}{2} [-2(2 - (-8)) - (-3)(3 - (-1)) + 1(-24 - (-2))] \\ &= \frac{1}{2} [-2(10) + 3(4) + 1(-22)] \\ &= \frac{1}{2} [-20 + 12 - 22] = \frac{1}{2} [-30] = -15 \end{aligned} \]
Since area is always positive, Area = \( |-15| = 15 \) sq units.
Q2
Show that points \( A(a, b+c), B(b, c+a), C(c, a+b) \) are collinear.
Solution:
Points are collinear if the area of the triangle formed by them is zero.
\[ \Delta = \frac{1}{2} \begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{vmatrix} \]
Applying \( C_1 \to C_1 + C_2 \):
\[ \Delta = \frac{1}{2} \begin{vmatrix} a+b+c & b+c & 1 \\ a+b+c & c+a & 1 \\ a+b+c & a+b & 1 \end{vmatrix} \]
Taking \( (a+b+c) \) common from \( C_1 \):
\[ \Delta = \frac{1}{2} (a+b+c) \begin{vmatrix} 1 & b+c & 1 \\ 1 & c+a & 1 \\ 1 & a+b & 1 \end{vmatrix} \]
Since \( C_1 \) and \( C_3 \) are identical, the determinant is zero.
Therefore, the points are collinear.
Q3
Find values of \( k \) if area of triangle is 4 sq. units and vertices are
(i) \( (k, 0), (4, 0), (0, 2) \)
(ii) \( (-2, 0), (0, 4), (0, k) \)
(i) \( (k, 0), (4, 0), (0, 2) \)
(ii) \( (-2, 0), (0, 4), (0, k) \)
Solution:
Since area is 4 sq. units, \( \Delta = \pm 4 \).
(i) Vertices: \( (k, 0), (4, 0), (0, 2) \)
\[ \frac{1}{2} \begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = \pm 4 \]
Expanding along C2:
\( \frac{1}{2} [-2(k-4)] = \pm 4 \)
\( -(k-4) = \pm 4 \Rightarrow -k + 4 = \pm 4 \).
- Case 1: \( -k + 4 = 4 \Rightarrow k = 0 \)
- Case 2: \( -k + 4 = -4 \Rightarrow k = 8 \)
So, \( k = 0, 8 \).
(ii) Vertices: \( (-2, 0), (0, 4), (0, k) \)
\[ \frac{1}{2} \begin{vmatrix} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{vmatrix} = \pm 4 \]
Expanding along C1:
\( \frac{1}{2} [-2(4-k)] = \pm 4 \)
\( -(4-k) = \pm 4 \Rightarrow k - 4 = \pm 4 \).
- Case 1: \( k - 4 = 4 \Rightarrow k = 8 \)
- Case 2: \( k - 4 = -4 \Rightarrow k = 0 \)
So, \( k = 0, 8 \).
Q4
(i) Find equation of line joining (1, 2) and (3, 6) using determinants.
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.
Solution:
(i) Let \( P(x, y) \) be any point on the line. The area of triangle formed by \( (1, 2), (3, 6), (x, y) \) is zero.
\[ \frac{1}{2} \begin{vmatrix} 1 & 2 & 1 \\ 3 & 6 & 1 \\ x & y & 1 \end{vmatrix} = 0 \]
\( 1(6-y) - 2(3-x) + 1(3y-6x) = 0 \)
\( 6 - y - 6 + 2x + 3y - 6x = 0 \)
\( 2y - 4x = 0 \Rightarrow y = 2x \).
(ii) Similarly for \( (3, 1) \) and \( (9, 3) \):
\[ \frac{1}{2} \begin{vmatrix} 3 & 1 & 1 \\ 9 & 3 & 1 \\ x & y & 1 \end{vmatrix} = 0 \]
\( 3(3-y) - 1(9-x) + 1(9y-3x) = 0 \)
\( 9 - 3y - 9 + x + 9y - 3x = 0 \)
\( 6y - 2x = 0 \Rightarrow x = 3y \).
Q5
If area of triangle is 35 sq units with vertices \( (2, -6), (5, 4) \) and \( (k, 4) \). Then \( k \) is- 12
- -2
- -12, -2
- 12, -2
- 12
- -2
- -12, -2
- 12, -2
Answer: (D)
Solution:
\[ \frac{1}{2} \begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} = \pm 35 \]
Expanding along C2 or R1:
\( 2(4-4) - (-6)(5-k) + 1(20-4k) = \pm 70 \)
\( 0 + 6(5-k) + 20 - 4k = \pm 70 \)
\( 30 - 6k + 20 - 4k = \pm 70 \)
\( 50 - 10k = \pm 70 \)
- Case 1: \( 50 - 10k = 70 \Rightarrow -10k = 20 \Rightarrow k = -2 \)
- Case 2: \( 50 - 10k = -70 \Rightarrow -10k = -120 \Rightarrow k = 12 \)
Therefore, \( k = 12, -2 \).