ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants Ex 4.3

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants Ex 4.3

Question 1.
Using determinants, find the area of the triangle whose vertices are :
(i) (3, 8), (- 4 , 2), (5, 1) (NCERT)
(ii) (2, 7), (1, 1), (10, 8)
Solution:
(i) The area Δ of triangle is given by Δ = \(\frac{1}{2}\left|\begin{array}{rrr}
3 & 8 & 1 \\
-4 & 2 & 1 \\
5 & 1 & 1
\end{array}\right|\) ;
operate R₂ → R₂ – R₁,
R₃ → R₃ – R₁
= \(\frac{1}{2}\left|\begin{array}{rrr}
3 & 8 & 1 \\
-7 & -6 & 0 \\
2 & -7 & 0
\end{array}\right|\) ;
expanding along C₃
Δ = \(\frac{1}{2} \times 1 \times\left|\begin{array}{rr}
-7 & -6 \\
2 & -7
\end{array}\right|\)
= \(\frac{1}{2}\) (49 + 12)
= \(\frac{61}{2}\) sq. units

(ii) The area Δ of triangle is given by
Δ = \(\frac{1}{2}\left|\begin{array}{rrr}
2 & 7 & 1 \\
1 & 1 & 1 \\
10 & 8 & 1
\end{array}\right|\)
operate R₂ → R₂ – R₁,
R₃ → R₃ – R₁
= \(\frac{1}{2}\left|\begin{array}{rrr}
2 & 7 & 1 \\
-1 & -6 & 0 \\
8 & 1 & 0
\end{array}\right|\)
Expanding along C₃
= \(\frac{1}{2} \times 1 \times\left|\begin{array}{rr}
-1 & -6 \\
8 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) (- 1 + 48)
= \(\frac{47}{2}\) sq. units.

Question 2.
Show that the points (1, 0), (6, 0), (0, 0) are collinear.
Solution:
Let the given points be A (1, 0); B (6, 0) and C (0, 0)
∴ area of ΔABC = \(\frac{1}{2}\left|\begin{array}{lll}
1 & 0 & 1 \\
6 & 0 & 1 \\
0 & 0 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) × 0 [∵ C₂ is a zero column]
= 0
Hence the given points A (1, 0), B (6, 0) and C (0, 0) are collinear.

Question 3.
Using determinants, determine whether the points P (a, b + c), Q (b, c + a) and R (c, a + b) form a triangle or not. (NCERT)
Solution:
Here area of ∆ABC = \(\frac{1}{2}\left|\begin{array}{lll}
a & b+c & 1 \\
b & c+a & 1 \\
c & a+b & 1
\end{array}\right|\) ;
Operate C₂ → C₂ + C₁
= \(\frac{1}{2}\left|\begin{array}{ccc}
a & a+b+c & 1 \\
b & a+b+c & 1 \\
c & a+b+c & 1
\end{array}\right|\) ;
Taking (a + b + c) common from C₂
= \(\frac{1}{2}\) (a + b + c) \(\left|\begin{array}{lll}
a & 1 & 1 \\
b & 1 & 1 \\
c & 1 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) (a + b + c) × 0
[∵ C₂ and C₃ are identical]
∴ area of ∆ABC formed by these points is 0.
Hence the given points are collinear.
Thus, given points do not form a triangle.

Question 4.
Using determinants, find the value of λ so that the points (λ, 7), (1, – 5) and (- 4, 5) are collinear.
Solution:
Given points A (λ, 7) ; B (1, – 5) and C (- 4, 5) are collinear
iff area of ∆ABC = 0
iff \(\frac{1}{2}\left|\begin{array}{rrr}
\lambda & 7 & 1 \\
1 & -5 & 1 \\
-4 & 5 & 1
\end{array}\right|\) = 0
iff λ (- 5 – 5) – 7 (1 + 4) + 1 (5 – 20) = 0
iff – 10λ – 35 – 15 = 0
iff – 10λ – 50 = 0
iff λ = – 5.

Question 5.
Using determinants, find the equation of the line passing through the points (1, 2) and (3, 6).
Solution:
Let (x, y) be any point on line joining (1, 2) and (3, 6).
Thus (x, y), (1, 2), (3, 6) are collinear.
∴ \(\frac{1}{2}\left|\begin{array}{lll}
x & y & 1 \\
1 & 2 & 1 \\
3 & 6 & 1
\end{array}\right|\) = 0 ;
operate R₂ → R₂ – R₁,
R₃ → R₃ – R₁
⇒ \(\left|\begin{array}{ccc}
x & y & 1 \\
1-x & 2-y & 0 \\
3-x & 6-y & 0
\end{array}\right|\) = 0 ;
expanding along C₃
⇒ (1 – x) (6 – y) – (2 – y) (3 – x) = 0
⇒ 6 – y – 6x + xy – 6 + 2x + 3y – xy = 0
⇒ 2y – 4x = 0
⇒ y = 2x be the required eqn. of line joining (1, 2) and (3, 6).

Question 6.
Find the value(s) of k if the area of the triangle with vertices (k, 0) (4, 0) and (0, 2) is 4 sq. units. (NCERT)
Solution:
Given, area of the triangle with vertices are (k, 0) (4, 0) and (0, 2) is given by
\(\frac{1}{2}\left\|\begin{array}{lll}
k & 0 & 1 \\
4 & 0 & 1 \\
0 & 2 & 1
\end{array}\right\|\)
also area of triangle = 4 sq. units

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

⇒ \(\left|\begin{array}{lll}
k & 0 & 1 \\
4 & 0 & 1 \\
0 & 2 & 1
\end{array}\right|\) = ± 8 ;

operate R₂ → R₂ – R₁,
R₃ → R₃ – R₁

⇒ \(\left|\begin{array}{ccc}
k & 0 & 1 \\
4-k & 0 & 0 \\
-k & 2 & 0
\end{array}\right|\) = ± 8 ;

expanding along C₃

⇒ \(\left|\begin{array}{cc}
4-k & 0 \\
-k & 2
\end{array}\right|\) = ± 8
⇒ 2 (4 – k) = ± 8
⇒ 4 – k = ± 4
⇒ k = 4 ± 4 = 8, 0.

Question 6 (Old).
If the area of a triangle with vertices (- 3, 0), (3, 0) and (0, k) is 9 sq. units, find the value(s) of k. (NCERT Exemplar)
Solution:
area of triangle = \(\frac{1}{2}\left|\begin{array}{rrr}
-3 & 0 & 1 \\
3 & 0 & 1 \\
0 & k & 1
\end{array}\right|\)
expanding along R₁
= |\(\frac{1}{2}\) [- 3 (0 – k) + 0 + 1 (3k)]|
= | 3k | sq. units
also given area of triangle = 9 sq. units
∴ | 3k | = 9
k = ± 3.

Question 7.
Find the value(s) of x if the area of the triangle with vertices (x, 4), (2, – 6) and (5, 4) is 35 sq. units. (NCERT)
Solution:
area Δ of triangle is given by Δ = \(\frac{1}{2}\left|\begin{array}{rrr}
x & 4 & 1 \\
2 & -6 & 1 \\
5 & 4 & 1
\end{array}\right|\)
operate R₂ → R₂ – R₁
R₃ → R₃ – R₁
= \(\frac{1}{2}\left|\begin{array}{ccc}
x & 4 & 1 \\
2-x & -10 & 0 \\
5-x & 0 & 0
\end{array}\right|\) ;
expanding along C₃
= \(\frac{1}{2}\left|\begin{array}{rr}
2-x & -10 \\
5-x & 0
\end{array}\right|\)
= \(\frac{1}{2}\) [10 (5 – x)]
= |5 (5 – x)| cm²
also given area of triangle = 35 cm²
∴ 5 (5 – x) = ± 35.
⇒ 5 – x = ± 7
⇒ x = 5 ± 7
= 12, – 2.

Question 8.
If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab.
Solution:
Since the given points (a, 0), (0, b) and (1, 1) are collinear.
Thus area of triangle formed for by given points is 0.
∴ ∆ = 0
⇒ \(\frac{1}{2}\left|\begin{array}{lll}
a & 0 & 1 \\
0 & b & 1 \\
1 & 1 & 1
\end{array}\right|\) = 0
operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁
\(\left|\begin{array}{ccc}
a & 0 & 1 \\
-a & b & 0 \\
1-a & 1 & 0
\end{array}\right|\) = 0 ;
expanding along C₃
⇒ \(\left|\begin{array}{cc}
-a & b \\
1-a & 1
\end{array}\right|\) = 0
⇒ – a – b (1 – a) = 0
⇒ – a – b + ab = 0
⇒ a + b = ab.