OP Malhotra Class 12 Maths Solutions Chapter 5 Determinants Ex 5(c)

Continuous practice using OP Malhotra Class 12 Solutions Chapter 5 Determinants Ex 5(c) can lead to a stronger grasp of mathematical concepts.

S Chand Class 12 ICSE Maths Solutions Chapter 5 Determinants Ex 5(c)

Question 1.
Find the area of the triangle whose vertices are :
(i) (- 8, – 2), (- 4, – 6), (- 1, 5)
(ii) (- 3, 5), (3, – 6), (7, 2)
Solution:
(i) required area of triangle
∆ = \(\left|\frac{1}{2}\left|\begin{array}{lll}
-8 & -2 & 1 \\
-4 & -6 & 1 \\
-1 & 5 & 1
\end{array}\right| \right|\)
expanding along R₁
= |\(\frac { 1 }{ 2 }\){(- 8(-6-5) + 2(- 4 + 1)+1 (-20-6)}|
= |\(\frac { 1 }{ 2 }\)[88-6-26]|
= 28 square units

(ii) required area of triangle
∆ = \(\left|\frac{1}{2}\left|\begin{array}{ccc}
-3 & 5 & 1 \\
3 & -6 & 1 \\
7 & 2 & 1
\end{array}\right| \right|\)
expanding along R₃
= |\(\frac { 1 }{ 2 }\)[1(6+42)-1(6-35)+1(18-15)]|
= |\(\frac { 1 }{ 2 }\)(48+41+3)|
= 46 square units

Question 2.
Using determinants, prove that the following points are eollinear
(i) (11, 7), (5, 5), (- 1, 3)
(ii) (0, 3), (4, 6), (- 8, – 3)
Solution:
(i) Here area of triangle formed given points is given by
∆ = \(\left|\frac{1}{2}\left|\begin{array}{ccc}
11 & 7 & 1 \\
5 & 5 & 1 \\
-1 & 3 & 1
\end{array}\right| \right|\)
= |\(\frac { 1 }{ 2 }\)[1(15+5)-1(33+7)+1(55-35)|
expanding along C₃
= |\(\frac { 1 }{ 2 }\)(20-40+20)| = 0 square units
Here the given points are collinear.

(ii) Here, area pf triangle formed by given points as vertices is given by
∆ = \(\left|\frac{1}{2}\left|\begin{array}{ccc}
0 & 3 & 1 \\
4 & 6 & 1 \\
-8 & -3 & 1
\end{array}\right| \right|\)
expanding along C₃
= \(\left|\frac{1}{2}[1(-12+48)-1(0+24)+1(0-12)]\right|\)
= \(\left|\frac{1}{2}(36-24-12)\right|\) = 0 square units
Here the given points are collinear.

Question 3.
Show that the points (b , c + a), (c, a + b) and (a, b + c) are collinear.
Solution:
Here, area of triangle formed by veritces as given points is given by

Thus, the given points are collinear.

Question 4.
Find x so that the points (3, – 2), (x, 2) and (8, 8) be on a line.
Solution:
Since, the given points (3, – 2), (x, 2) and (8, 8) lies on a line
∴ given points are collinear. Thus the area of triangle formed by given points as vertices be zero.
\(\left|\frac{1}{2}\left|\begin{array}{ccc}
3 & -2 & 1 \\
x & 2 & 1 \\
8 & 8 & 1
\end{array}\right| \right|\) = 0;
expanding along C₃
\(|1(8 x-16)-1(24+16)+1(6+2 x)|=0\) = 0
⇒ |(8x – 16 – 40 + 6 + 2x)| = 0
⇒ |10x – 50| = 0
⇒ 10x – 50 = 0
⇒ x = 5

Question 5.
If (x, y), (a, 0), (0, b) are collinear, then using determinants prove that \(\frac{x}{a}+\frac{y}{b}\) = 1.
Solution:
Since given points (x, y), (a, 0), (0, b) are collinear
∴ area area of ∆ formed by given points as vertices be equal to 0.
∴ \(\left|\frac{1}{2}\left|\begin{array}{lll}
x & y & 1 \\
a & 0 & 1 \\
0 & b & 1
\end{array}\right| \right|\) = 0;
expanding along C₃
|1 (ab – 0) – 1(bx – 0) + 1(0-ay)| = 0
⇒ ab – bx – ay = 0
⇒ bx + ay = ab
⇒ \(\frac{x}{a}+\frac{y}{b}\) = 1 which is the required result