GSEB Solutions Class 12 Maths Chapter 4 Determinants Ex 4.3

   

Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 4 Determinants Ex 4.3 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 4 Determinants Ex 4.3

GSEB Solutions Class 12 Maths Chapter 4 Determinants Ex 4.3

Question 1.
Find the area of the triangle with vertices at the points 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 A of the ∆ whose vertices are (1, 0), (6, 0), (4, 3) is given by
A = \(\frac { 1 }{ 2 }\)\(\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|\) = \(\frac { 1 }{ 2 }\)\(\left|\begin{array}{lll} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{array}\right|\)
Expanding with the help of elements of the second column, we get
A = \(\frac { 1 }{ 2 }\)| 3 x (1 – 6) | = \(\frac { 15 }{ 2 }\) = 7 \(\frac { 1 }{ 2 }\) Sq. units.

(ii) The vertices of ∆ ABC are A (2, 7), B (1, 1), C (10, 8).
The area A of ∆ ABC = \(\frac { 1 }{ 2 }\)\(\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|\) = \(\frac { 1 }{ 2 }\)\(\left|\begin{array}{lll} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{array}\right|\)
Expanding with the help of elements of I column, we pet
A = \(\frac { 1 }{ 2 }\)[2(1 – 8) – 1(7 – 8) + 10(7 – 1)]
= \(\frac { 1 }{ 2 }\) [- 14 + 1 + 60] = \(\frac { 47 }{ 2 }\) = 23\(\frac { 1 }{ 2 }\) Sq. units.

(iii) The vertices of ∆ ABC are A (-2, -3), B (3, 2), C (-1, -8).
Area of ∆ ABC = – \(\frac { 1 }{ 2 }\)\(\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|\) = \(\left|\begin{array}{ccc} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \end{array}\right|\)
= \(\frac { 1 }{ 2 }\)[-2(2 + 8) + 3(3 + 1) + 1(- 24 + 2)]
= \(\frac { 1 }{ 2 }\)(- 20 + 12 – 22)
= (- 30) = – 15.
∴ Ignoring – ve sign,
Area of ∆ ABC = 15 sq. units.

GSEB Solutions Class 12 Maths Chapter 4 Determinants Ex 4.3

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), C (c, a + b)
GSEB Solutions Class 12 Maths Chapter 4 Determinants Ex 4.3 1
Hence, the points A, B and C 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), (1, 4), (0, k)
Solution:
(i) Area A of triangle with vertices (k, 0), (4, 0), (0, 2)
= \(\frac { 1 }{ 2 }\)\(\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|\) = \(\frac { 1 }{ 2 }\)\(\left|\begin{array}{lll}
k & 0 & 1 \\
4 & 0 & 1 \\
0 & 2 & 1
\end{array}\right|\)
Expanding with the help of elements of II column, we get
= \(\frac { 1 }{ 2 }\) [- 2(k – 4)] = – (k – 4) = ± 4 (Given)
Taking + ve sign, – (k – 4) = 4 ∴ k = 0.
Taking – ve sign, – (k – 4) = – 4 ∴ k = 8.

(ii) The area A of the triangle whose vertices are (- 2, 0), (0, 4), (0, k)
= \(\frac { 1 }{ 2 }\)\(\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|\) = \(\left|\begin{array}{ccc} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{array}\right|\)
= \(\frac { 1 }{ 2 }\) x (- 2) (4 – k) = k – 4 = ± 4 (Given)
Taking +ve sign, k – 4 = 4 ∴ k = 8.
Taking -ve sign, k – 4 = – 4 ∴ k = 0.

GSEB Solutions Class 12 Maths Chapter 4 Determinants Ex 4.3

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) and (9, 3), using determinants.
Solution:
(i) Let (x, y) be the third points on the line joining (1, 2) and (3, 6).
The area A of triangle with vertices (x, y), (1, 2) and (3, 6)
= \(\frac { 1 }{ 2 }\)\(\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|\) = \(\left|\begin{array}{lll} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{array}\right|\)
Since the three points are coliinear,
∴ \(\left|\begin{array}{lll} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{array}\right|\) = 0.
or x(2 – 6) – y(1 – 3) + 1(6 – 6) = 0
or – 4x + 2y = 0
or 2x – y = 0.
It is the required equation of the line.

(ii) Let (x, y) be the third point on the line joining (3, 1) and (9, 3).
The area A of the triangle whose vertices are (x, y), (3, 1) and (9, 3)
= \(\frac { 1 }{ 2 }\)\(\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|\) = \(\left|\begin{array}{lll} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{array}\right|\)
= \(\frac { 1 }{ 2 }\)[x(1 – 3)-y(3-9)+ 1(9-9)]
= \(\frac { 1 }{ 2 }\)[x(-2)-y(-6)].
As these points are coliinear, the area of the triangle is zero.
∴ x(- 2) – y(- 6) = 0
or 2x – 6y = 0
or x-3y = 0.
It is the required equation of the line.

GSEB Solutions Class 12 Maths Chapter 4 Determinants Ex 4.3

Question 5.
If the area of a triangle with vertices (2, – 6), (5, 4) and (k, 4) is 35 sq. units, then k is
(A) 12
(B) – 2
(C) – 12, – 2
(D) 12,- 2
Solution:
The area of the triangle with the given vertices
= \(\frac { 1 }{ 2 }\)\(\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|\) = \(\left|\begin{array}{lll} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{array}\right|\)
= \(\frac { 1 }{ 2 }\)[2(4 – 4) + 6(5 -k) + 1(20 – 4k) ]
= \(\frac { 1 }{ 2 }\)[(30 – 6k) + 20 – 4k] = \(\frac { 1 }{ 2 }\) [50 – 10k].
The area = 35 sq. units. ∴ \(\frac { 1 }{ 2 }\) [50 – 10k] = ± 35
Taking +ve sign, \(\frac { 1 }{ 2 }\)[50 – 10k] = 35
or 50 – 10k = 70
or 10k = 50 – 70 = – 20
∴ k = -2.
Taking – ve sign, \(\frac { 1 }{ 2 }\)[50 – 10k] = – 35
or 50 – 10k = – 70
or 50 + 70 = 10k
∴ k = 12.
Hence, k = 12, – 2.
Part (D) is the required answer.

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