Anand Classes brings you the most accurate and student-friendly NCERT Solutions for Determinants Exercise 4.2 – Class 12 Chapter 4, designed to help you strengthen your conceptual understanding and score full marks in board exams. These solutions follow the latest CBSE pattern and provide step-by-step explanations for every question, making them ideal for quick revision and doubt-clearing. Whether you’re preparing for board exams or competitive entrances, these well-structured notes will improve accuracy and boost confidence. Click the print button to download study material and notes.
Access exam-oriented NCERT Solutions for Chapter 4 Determinants Exercise 4.2 of Class 12 Maths
NCERT Question 1 : Find the area of the triangle with vertices at the points:
(i) $(1,0),(6,0),(4,3)$
(ii) $(2,7),(1,1),(10,8)$
(iii) $(-2,-3),(3,2),(-1,-8)$
Solution
The area of a triangle with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ can be found using the determinant formula:
$$
\text{Area} = \frac{1}{2}
\begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{vmatrix}.
$$
(i) Vertices: $(1,0),(6,0),(4,3)$
$$
\text{Area}=\frac{1}{2}\begin{vmatrix}1 & 0 & 1\\6 & 0 & 1\\4 & 3 & 1
\end{vmatrix}$$
$$\text{Area}=\frac{1}{2}\bigl[1(0-3)-0(6-4)+1(18-0)\bigr]$$
$$\text{Area}=\frac{1}{2}\bigl(-3+18\bigr)=\frac{15}{2}$$
Answer (i):
$$
\boxed{\frac{15}{2}\text{ square units}}
$$
(ii) Vertices: $(2,7),(1,1),(10,8)$
$$
\text{Area}=\frac{1}{2}\begin{vmatrix}
2 & 7 & 1\\
1 & 1 & 1\\
10 & 8 & 1
\end{vmatrix}$$
$$\text{Area}=\frac{1}{2}\bigl[2(1-8)-7(1-10)+1(8-10)\bigr]$$
$$\text{Area}=\frac{1}{2}\bigl(-14+63-2\bigr)=\frac{47}{2}$$
Answer (ii):
$$
\boxed{\frac{47}{2}\text{ square units}}
$$
(iii) Vertices: $(-2,-3),(3,2),(-1,-8)$
$$
\text{Area}=\frac{1}{2}\begin{vmatrix}
-2 & -3 & 1\\
3 & 2 & 1\\
-1 & -8 & 1
\end{vmatrix}$$
$$\text{Area}=\frac{1}{2}\bigl[-2(10)+3(4)-22\bigr]$$
$$\text{Area}=\frac{1}{2}\bigl(-20+12-22\bigr)=15$$
Answer (iii):
$$
\boxed{15\text{ square units}}
$$
Finding the area of a triangle using determinants is a powerful application of determinants in coordinate geometry. Students can calculate the area efficiently and accurately. Practicing such determinant-based area problems helps strengthen spatial reasoning and algebraic manipulation skills for NCERT Class 12 Mathematics, especially useful for exams and competitive tests where applications of determinants are tested.
NCERT Question 2 : Show that the points
$A(a,b+c),B(b,c+a),C(c,a+b)$ are collinear.
Solution
Points are collinear iff the area of the triangle they form is zero. Using the determinant formula for area,
$$
\text{Area}=\frac{1}{2}
\begin{vmatrix}
a & b+c & 1\\
b & c+a & 1\\
c & a+b & 1
\end{vmatrix}.
$$
Let
$$D=
\begin{vmatrix}
a & b+c & 1\\
b & c+a & 1\\
c & a+b & 1
\end{vmatrix}.
$$
Expand $D$ along the first row:
$$
D= a\begin{vmatrix}c+a & 1\\ a+b & 1\end{vmatrix}-(b+c)\begin{vmatrix}b & 1\\ c & 1\end{vmatrix}+1\begin{vmatrix}b & c+a\\ c & a+b\end{vmatrix}
$$
Compute each $2\times2$ minor:
$$
\begin{vmatrix}c+a & 1\\ a+b & 1\end{vmatrix}= (c+a)\cdot 1 – 1\cdot (a+b) = c – b$$
$$\begin{vmatrix}b & 1\\ c & 1\end{vmatrix}= b\cdot 1 – 1\cdot c = b – c$$
$$\begin{vmatrix}b & c+a\\ c & a+b\end{vmatrix}= b(a+b) – (c+a)c = ab + b^{2} – c^{2} – ac$$
Substitute back into $D$:
$$
\begin{aligned}
D
&= a(c-b) – (b+c)(b-c) + \bigl(ab + b^{2} – c^{2} – ac\bigr).
\end{aligned}
$$
Simplify step by step. Note $(b+c)(b-c)=b^{2}-c^{2}$, so
$$
D= ac – ab – (b^{2}-c^{2}) + ab + b^{2} – c^{2} – ac$$
$$D= \bigl(ac – ab – b^{2} + c^{2} + ab + b^{2} – c^{2} – ac\bigr) = 0$$
Thus $\text{Area}=\dfrac{1}{2}D=\dfrac{1}{2}\cdot 0=0$, so the three points are collinear.
Final Result
$$\boxed{\text{The points }A(a,b+c),B(b,c+a),C(c,a+b)\text{ are collinear.}}$$
This determinant-based proof shows a neat coordinate geometry technique from NCERT Class 12: using the $3times 3$ area determinant to test collinearity. Understanding how to set up and simplify such determinants helps students quickly verify geometric properties of points, strengthens algebraic manipulation skills, and is invaluable for NCERT exam practice and competitive maths preparation.
NCERT Question 3 : Find the values of $k$ if the area of the triangle is $4$ square units for the vertices
(i) $(k,0),\ (4,0),\ (0,2)$
(ii) $(-2,0),\ (0,4),\ (0,k)$
Solution (i)
Area formula using a $3\times 3$ determinant:
$$
\text{Area}=\frac{1}{2}
\begin{vmatrix}
x_1 & y_1 & 1\\
x_2 & y_2 & 1\\
x_3 & y_3 & 1
\end{vmatrix}.
$$
For $(k,0),(4,0),(0,2)$ the determinant is
$$
D =
\begin{vmatrix}
k & 0 & 1\\
4 & 0 & 1\\
0 & 2 & 1
\end{vmatrix}$$
$$D= k(0-2)-0(4-0)+1(8-0)= -2k+8$$
Since $\text{Area}=\dfrac{1}{2}|D|=4$ we have $|D|=8$, so
$$
-2k+8 = \pm 8.
$$
Solve both cases:
- $-2k+8=8 \implies -2k=0 \implies k=0.$
- $-2k+8=-8 \implies -2k=-16 \implies k=8.$
Therefore for (i)
$$\boxed{k=0\ \text{or}\ k=8.}$$
(ii)
For $(-2,0),(0,4),(0,k)$ compute the determinant:
$$
D =\begin{vmatrix}
-2 & 0 & 1\\
0 & 4 & 1\\
0 & k & 1
\end{vmatrix}$$
$$D= -2(4-k)-0+1(0-0)= -8+2k$$
Again $\dfrac{1}{2}|D|=4\implies |D|=8$, so
$$
-8+2k=\pm 8.
$$
Solve both cases:
- $-8+2k=8 \implies 2k=16 \implies k=8.$
- $-8+2k=-8 \implies 2k=0 \implies k=0.$
Therefore for (ii)
$$\boxed{k=0\ \text{or}\ k=8.}$$
Final Answer
$$\boxed{\text{For both (i) and (ii): }k=0\ \text{or}\ k=8.}$$
Determining values of parameters from triangle area using the determinant formula is a practical NCERT Class 12 application—transforming vertex coordinates into a $3times 3$ determinant and solving $ tfrac{1}{2}|D|=text{given area}$ gives the required parameter values quickly and reliably. Practising these coordinate-geometry determinant problems improves speed and accuracy for board exams and competitive tests.
NCERT Question 4 :
(i) Find equation of the line joining $(1,2)$ and $(3,6)$ using determinants.
(ii) Find equation of the line joining $(3,1)$ and $(9,3)$ using determinants.
Solution :
Let us assume, $A(x, y)$ be any vertex of a triangle.
(i)
All three points $(1,2),(3,6),(x,y)$ are collinear iff the area of the triangle they form is zero. Using the $3\times 3$ determinant form for area,
$$
\frac{1}{2}
\begin{vmatrix}
1 & 2 & 1\\
3 & 6 & 1\\
x & y & 1
\end{vmatrix}=0.
$$
Thus the determinant must vanish:
$$
\begin{vmatrix}
1 & 2 & 1\\
3 & 6 & 1\\
x & y & 1
\end{vmatrix}=0.
$$
Expand the determinant (expand along the third row or compute directly):
$$
x(2-6)-y(1-3)+1(6-6)=0.
$$
Simplify:
$$
-4x+2y+0=0\quad\Longrightarrow\quad 2y-4x=0.
$$
Divide by $2$:
$$
y-2x=0\quad\Longrightarrow\quad y=2x.
$$
Which is equation of a line.
Final Result (i)
$$\boxed{y=2x}$$
(ii)
Same approach for points $(3,1),(9,3),(x,y)$:
$$
\frac{1}{2}
\begin{vmatrix}
3 & 1 & 1\\
9 & 3 & 1\\
x & y & 1
\end{vmatrix}=0$$
$$\begin{vmatrix}
3 & 1 & 1\\
9 & 3 & 1\\
x & y & 1
\end{vmatrix}=0.
$$
Expanding gives
$$
x(1-3)-y(3-9)+1(9-9)=0.
$$
Simplify:
$$
-2x-(-6y)+0=0\quad\Longrightarrow\quad -2x+6y=0.
$$
Divide by $2$:
$$
-x+3y=0\quad\Longrightarrow\quad x-3y=0.
$$
Which is equation of a line.
Final Result (ii)
$$\boxed{x-3y=0}$$
Using determinants to derive the equation of a line (by setting the area determinant to zero) is a compact and reliable NCERT Class 12 technique—apply the $3times 3$ determinant with coordinates $(x_i,y_i,1)$ to test collinearity and obtain line equations such as $y=2x$ and $x-3y=0$. Practising this determinant method strengthens coordinate-geometry skills and speeds up solving line and collinearity problems for board exams and competitive tests.
NCERT Question 5 : If area of triangle is 35 square units with vertices $(2,-6),\ (5,4)\ \text{and}\ (k,4)$
Then $k$ is
$(A) 12 \ (B) –2 \ (C) –12, –2 \ (D) 12, –2 $
Solution
Use the triangle area formula in terms of a determinant:
$$
\text{Area} = \frac{1}{2}
\begin{vmatrix}
x_1 & y_1 & 1\\
x_2 & y_2 & 1\\
x_3 & y_3 & 1
\end{vmatrix}.
$$
Substitute $(x_1,y_1) = (2,-6),\ (x_2,y_2) = (5,4),\ (x_3,y_3) = (k,4)$:
$$
\text{Area}=\frac{1}{2}\begin{vmatrix}
2 & -6 & 1\\
5 & 4 & 1\\
k & 4 & 1
\end{vmatrix}
=35.
$$
Compute the determinant:
$$
D=\begin{vmatrix}
2 & -6 & 1\\
5 & 4 & 1\\
k & 4 & 1
\end{vmatrix}$$
$$D=2(4-4)-(-6)(5-k)+1(20-4k)$$
$$D=0-(-6)(5-k)+(20-4k)$$
$$D=6(5-k)+20-4k$$
$$D=30-6k+20-4k=50-10k$$
So,
$$
\frac{1}{2}|50-10k|=35\quad\Longrightarrow\quad |50-10k|=70.
$$
Now solve the absolute value equation:
- $50-10k=70\implies -10k=20\implies k=-2.$
- $50-10k=-70\implies -10k=-120\implies k=12.$
Final Answer: Option : D
$$\boxed{k=-2\ \text{or}\ k=12.}$$
This problem uses the determinant formula for the area of a triangle to relate coordinates and the unknown parameter $k$. By expressing the area and solving the resulting algebraic equation, we find the possible values of $k$. Understanding such coordinate geometry applications of determinants enhances problem-solving skills for NCERT Class 12 Mathematics and prepares students for board exams and competitive tests.
