Exercise 4.2 (Revised) - Chapter 4 Determinants class 12 ncert solutions Maths - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Chapter 4 - Determinants NCERT Solutions Class 12 Maths | Step-by-Step Solutions & Explanations

Ex 4.2 Question 1:
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) $(-2,-3),(3,2),(-1,-8)$

Answer
(i) The area of the triangle with vertices $(1,0),(6,0),(4,3)$ is given by the relation,
$
\begin{aligned}
\Delta & =\frac{1}{2}\left|\begin{array}{lll}
1 & 0 & 1 \\
6 & 0 & 1 \\
4 & 3 & 1
\end{array}\right| \\
& =\frac{1}{2}[1(0-3)-0(6-4)+1(18-0)] \\
& =\frac{1}{2}[-3+18]=\frac{15}{2} \text { square units }
\end{aligned}
$

(ii) The area of the triangle with vertices $(2,7),(1,1),(10,8)$ is given by the relation,
$
\begin{aligned}
\Delta & =\frac{1}{2}\left|\begin{array}{ccc}
2 & 7 & 1 \\
1 & 1 & 1 \\
10 & 8 & 1
\end{array}\right| \\
& =\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{1}{2}[-16+63] \\
& =\frac{47}{2} \text { square units }
\end{aligned}
$

(iii) The area of the triangle with vertices $(-2,-3),(3,2),(-1,-8)$ is given by the relation,
$
\begin{aligned}
\Delta & =\frac{1}{2}\left|\begin{array}{rrr}
-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}[-2(10)+3(4)+1(-22)] \\
& =\frac{1}{2}[-20+12-22] \\
& =-\frac{30}{2}=-15
\end{aligned}
$

Hence, the area of the triangle is $|-15|=15$ square units

Ex 4.2 Question 2:
Show that points
$\mathrm{A}(a, b+c), \mathrm{B}(b, c+a), \mathrm{C}(c, a+b)$ are collinear
Answer
Area of $\triangle A B C$ is given by the relation,
$
\begin{aligned}
\Delta & =\frac{1}{2}\left|\begin{array}{lll}
a & b+c & 1 \\
b & c+a & 1 \\
c & a+b & 1
\end{array}\right| \\
& \left.=\frac{1}{2}\left|\begin{array}{ccc}
a & b+c & 1 \\
b-a & a-b & 0 \\
c-a & a-c & 0
\end{array}\right| \text { (Applying } \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1 \text { and } \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1\right) \\
& =\frac{1}{2}(a-b)(c-a)\left|\begin{array}{ccc}
a & b+c & 1 \\
-1 & 1 & 0 \\
1 & -1 & 0
\end{array}\right| \\
& =\frac{1}{2}(a-b)(c-a)\left|\begin{array}{ccc}
a & b+c & 1 \\
-1 & 1 & 0 \\
0 & 0 & 0
\end{array}\right|\left(\text { Applying } \mathrm{R}_3 \rightarrow \mathrm{R}_3+\mathrm{R}_2\right) \\
& =0 \quad\left(\text { All elements of } \mathrm{R}_3 \text { are } 0\right)
\end{aligned}
$

Thus, the area of the triangle formed by points $\mathrm{A}, \mathrm{B}$, and $\mathrm{C}$ is zero. Hence, the points A, B, and C are collinear.

Ex 4.2 Question 3:
Find values of $k$ if area of triangle is 4 square units and vertices are
(i) $(k, 0),(4,0),(0,2)$ (ii) $(-2,0),(0,4),(0, k)$

Answer
We know that the area of a triangle whose vertices are $\left(x_1, y_1\right),\left(x_2, y_2\right)$, and $\left(x_3, y_3\right)$ is the absolute value of the determinant $(\Delta)$, where
$
\Delta=\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|
$

It is given that the area of triangle is 4 square units.
$
\therefore \Delta= \pm 4 \text {. }
$
(i) The area of the triangle with vertices $(k, 0),(4,0),(0,2)$ is given by the relation,
$
\begin{aligned}
& \Delta=\frac{1}{2}\left|\begin{array}{lll}
k & 0 & 1 \\
4 & 0 & 1 \\
0 & 2 & 1
\end{array}\right| \\
& =\frac{1}{2}[k(0-2)-0(4-0)+1(8-0)] \\
& =\frac{1}{2}[-2 k+8]=-k+4 \\
& \therefore-k+4= \pm 4
\end{aligned}
$

$
\begin{aligned}
& \text { When }-k+4=-4, k=8 . \\
& \text { When }-k+4=4, k=0 . \\
& \text { Hence, } k=0,8 \text {. }
\end{aligned}
$
(ii) The area of the triangle with vertices $(-2,0),(0,4),(0, k)$ is given by the relation,
$
\begin{aligned}
& \Delta=\frac{1}{2}\left|\begin{array}{ccc}
-2 & 0 & 1 \\
0 & 4 & 1 \\
0 & k & 1
\end{array}\right| \\
& =\frac{1}{2}[-2(4-k)] \\
& =k-4
\end{aligned}
$
$
\therefore k-4= \pm 4
$
$
\begin{aligned}
& \text { When } k-4=-4, k=0 \text {. } \\
& \text { When } k-4=4, k=8 \text {. } \\
& \text { Hence, } k=0,8 \text {. }
\end{aligned}
$

Ex 4.2 Question 4:
(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

Answer
(i) Let $\mathrm{P}(x, y)$ be any point on the line joining points $\mathrm{A}(1,2)$ and $\mathrm{B}(3,6)$. Then, the points $A, B$, and $P$ are collinear. Therefore, the area of triangle $A B P$ will be zero.
$
\begin{aligned}
& \therefore \frac{1}{2}\left|\begin{array}{lll}
1 & 2 & 1 \\
3 & 6 & 1 \\
x & y & 1
\end{array}\right|=0 \\
& \Rightarrow \frac{1}{2}[1(6-y)-2(3-x)+1(3 y-6 x)]=0 \\
& \Rightarrow 6-y-6+2 x+3 y-6 x=0 \\
& \Rightarrow 2 y-4 x=0 \\
& \Rightarrow y=2 x
\end{aligned}
$

Hence, the equation of the line joining the given points is $y=2 x$.
(ii) Let $\mathrm{P}(x, y)$ be any point on the line joining points $\mathrm{A}(3,1)$ and
$B(9,3)$. Then, the points $A, B$, and $P$ are collinear. Therefore, the area of triangle ABP will be zero.
$
\begin{aligned}
& \therefore \frac{1}{2}\left|\begin{array}{lll}
3 & 1 & 1 \\
9 & 3 & 1 \\
x & y & 1
\end{array}\right|=0 \\
& \Rightarrow \frac{1}{2}[3(3-y)-1(9-x)+1(9 y-3 x)]=0 \\
& \Rightarrow 9-3 y-9+x+9 y-3 x=0 \\
& \Rightarrow 6 y-2 x=0 \\
& \Rightarrow x-3 y=0
\end{aligned}
$

Hence, the equation of the line joining the given points is $x-3 y=0$.

Ex 4.2 Question 5:
If area of triangle is 35 square units with vertices $(2,-6),(5,4)$, and $(k, 4)$. Then $k$ is
A. 12
B. -2
C. $-12,-2$
D. $12,-2$

Answer
Answer: D
The area of the triangle with vertices $(2,-6),(5,4)$, and $(k, 4)$ is given by the relation,
$
\begin{aligned}
\Delta & =\frac{1}{2}\left|\begin{array}{ccc}
2 & -6 & 1 \\
5 & 4 & 1 \\
k & 4 & 1
\end{array}\right| \\
& =\frac{1}{2}[2(4-4)+6(5-k)+1(20-4 k)] \\
& =\frac{1}{2}[30-6 k+20-4 k] \\
& =\frac{1}{2}[50-10 k] \\
& =25-5 k
\end{aligned}
$

It is given that the area of the triangle is $\pm 35$.
Therefore, we have:
$
\begin{aligned}
& \Rightarrow 25-5 k= \pm 35 \\
& \Rightarrow 5(5-k)= \pm 35 \\
& \Rightarrow 5-k= \pm 7
\end{aligned}
$

When $5-k=-7, k=5+7=12$.
When $5-k=7, k=5-7=-2$.
Hence, $k=12,-2$.
The correct answer is D.