CBSE Class 12 Maths Chapter 4 Determinants – NCERT Solutions 2025–26

Exercise 4.2

1. Find the area of the triangle with vertices at the point given in each of the following:

i. \[\left( \text{1,0} \right)\text{,}\left( \text{6,0} \right)\text{,}\left( \text{4,3} \right)\]

Ans: Given vertices,\[\left( \text{1,0} \right)\text{,}\left( \text{6,0} \right)\text{,}\left( \text{4,3} \right)\]

Thus, the area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix}   \text{1} & \text{0} & \text{1}  \\   \text{6} & \text{0} & \text{1}  \\   \text{4} & \text{3} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \Delta =\dfrac{\text{1}}{\text{2}}\left[ \text{1}\left( \text{0-3} \right)\text{-0}\left( \text{6-4} \right)\text{+1}\left( \text{18-0} \right) \right]\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{-3+18} \right]\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{15}}{\text{2}}\] square units

$\therefore $Area of the triangle with vertices \[\left( \text{1,0} \right)\text{,}\left( \text{6,0} \right)\text{,}\left( \text{4,3} \right)\] is \[\dfrac{\text{15}}{\text{2}}\] square units.

ii. \[\left( \text{2,7} \right)\text{,}\left( \text{1,1} \right)\text{,}\left( \text{10,8} \right)\]

Ans: Given vertices,\[\left( \text{2,7} \right)\text{,}\left( \text{1,1} \right)\text{,}\left( \text{10,8} \right)\]

The area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{2} & \text{7} & \text{1}  \\  \text{1} & \text{1} & \text{1}  \\   \text{10} & \text{8} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{2}\left( \text{1-8} \right)\text{-7}\left( \text{1-10} \right)\text{+1}\left( \text{8-10} \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{2}\left( \text{-7} \right)\text{-7}\left( \text{-9} \right)\text{+1}\left( \text{-2} \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{-16+63} \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{47}}{\text{2}}\] square units

$\therefore $Area of the triangle with vertices \[\left( \text{2,7} \right)\text{,}\left( \text{1,1} \right)\text{,}\left( \text{10,8} \right)\] is \[\dfrac{47}{\text{2}}\] square units.

iii. \[\text{(-2,-3),}\left( \text{3,2} \right)\text{,(-1,-8)}\]

Ans: Given vertices,\[\text{(-2,-3),}\left( \text{3,2} \right)\text{,(-1,-8)}\]

The area of the triangle with vertices \[\text{(-2,-3),}\left( \text{3,2} \right)\text{,(-1,-8)}\] is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{-2} & \text{-3} & \text{1}  \\  \text{3} & \text{2} & \text{1}  \\  \text{-1} & \text{-8} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{-2}\left( \text{2+8} \right)\text{+3}\left( \text{3+1} \right)\text{+1}\left( \text{-24+2} \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{-20+12-22} \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ }=-\dfrac{\text{30}}{\text{2}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =-15}\]

$\therefore $The area of the triangle with vertices \[\left( \text{2,7} \right)\text{,}\left( \text{1,1} \right)\text{,}\left( \text{10,8} \right)\]is \[\left| \text{-15} \right|\text{=15}\] square units.

2. Show that points \[\text{A}\left( \text{a,b+c} \right)\text{,B}\left( \text{b,c +a} \right)\text{,C}\left( \text{c,a +b} \right)\] are collinear.

Ans: To show that the points \[\text{A}\left( \text{a,b+c} \right)\text{,B}\left( \text{b,c +a} \right)\text{, C}\left( \text{c, a +b} \right)\] are collinear, the area of the triangle formed by these points as vertices should be zero.

$\therefore $ Area of \[\text{ }\!\!\Delta\!\!\text{ ABC}\] is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix}   \text{a} & \text{b+c} & \text{1}  \\   \text{b} & \text{c+a} & \text{1}  \\   \text{c} & \text{a+b} & \text{1}  \\ \end{matrix} \right|\]

Applying the row operations, \[R_2\to R_2-R_1\] and \[R_3\to R_3-R_1\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix}  \text{a} & \text{b+c} & \text{1}  \\ \text{b-a} & \text{a-b} & \text{0}  \\ \text{c-a} & \text{a-c} & \text{0}  \\ \end{matrix} \right|\]

Taking out $\left( a-b \right)$ and $\left( c-a \right)$ common from \[R_2\] and \[R_3\] respectively,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left( \text{a-b} \right)\left( \text{c-a} \right)\left| \begin{matrix} \text{a} & \text{b+c} & \text{1}  \\ \text{-1} & \text{1} & \text{0}  \\ \text{1} & \text{-1} & \text{0}  \\ \end{matrix} \right|\]

Applying the row operation \[R_3\to R_3\text{+}R_2\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left( \text{a-b} \right)\left( \text{c-a} \right)\left| \begin{matrix} \text{a} & \text{b+c} & \text{1}  \\   \text{-1} & \text{1} & \text{0}  \\  \text{0} & \text{0} & \text{0}  \\ \end{matrix} \right|\]

We know that when all the elements of a row or a column in a determinant are zero then the value of the determinant is always zero.

\[\therefore \Delta \text{=0}\]                                           

Thus, the area of the triangle formed by points \[\text{A}\] , \[\text{B}\] and \[\text{C}\] is zero.

Hence, the points \[\text{A}\left( \text{a,b+c} \right)\text{,B}\left( \text{b,c +a} \right)\text{,C}\left( \text{c,a +b} \right)\] are collinear.

3. Find values of \[\text{k}\] if area of triangle is \[\text{4}\] square units and vertices are:

i. \[\left( \text{k,0} \right)\text{,}\left( \text{4,0} \right)\text{,}\left( \text{0,2} \right)\]

Ans: Given vertices are \[\left( \text{k,0} \right)\text{,}\left( \text{4,0} \right)\text{,}\left( \text{0,2} \right)\].

We know, the area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix}  \text{k} & \text{0} & \text{1}  \\  \text{4} & \text{0} & \text{1}  \\  \text{0} & \text{2} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{k}\left( \text{0-2} \right)\text{-0}\left( \text{4-0} \right)\text{+1}\left( \text{8-0} \right) \right]\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{-2k+8} \right]\]

\[\therefore \Delta =-k+4\]

Since the area is given to be \[\text{4}\] square units, thus

$-k+4=\pm 4$

When \[-k+4=-4\]

\[\therefore k=8\].

When \[-k+4=4\]

\[\therefore k=0\].

Hence, \[\text{k=0,8}\].

ii. \[\text{(-2,0),}\left( \text{0,4} \right)\text{,}\left( \text{0,k} \right)\]

Ans: Given vertices are \[\text{(-2,0),}\left( \text{0,4} \right)\text{,}\left( \text{0,k} \right)\].

The area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix}  \text{-2} & \text{0} & \text{1}  \\  \text{0} & \text{4} & \text{1}  \\  \text{0} & \text{k} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ -2\left( 4-k \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ }=k-4\]

Since the area is given to be \[\text{4}\] square units, thus

\[k-4=\pm 4\]

When \[k-4=-4\]

\[\therefore k=0\].

When \[k-4=4\]

\[\therefore k=8\].

Hence, \[k=0,8\].

4. Determine the following:

i. Find the equation of line joining \[\left( \text{1,2} \right)\] and \[\left( \text{3,6} \right)\] using determinants.

Ans: Let us assume a point, \[\text{P}\left( \text{x, y} \right)\] on the line joining points \[\text{A}\left( \text{1,2} \right)\] and \[\text{B}\left( \text{3,6} \right)\] .

Then, the points \[\text{A}\],$B$ and $P$ are collinear.

Thus, the area of the triangle \[\text{ABP}\] will be zero.

\[\therefore \dfrac{\text{1}}{\text{2}}\left| \begin{matrix}  \text{1} & \text{2} & \text{1}  \\ \text{3} & \text{6} & \text{1}  \\  \text{x} & \text{y} & \text{1}  \\ \end{matrix} \right|\text{=0}\]

\[\Rightarrow \dfrac{\text{1}}{\text{2}}\left[ \text{1}\left( \text{6-y} \right)\text{-2}\left( \text{3-x} \right)\text{+1}\left( \text{3y-6x} \right) \right]\text{=0}\]

\[\Rightarrow \text{6-y-6+2x+3y-6x=0}\]

\[\Rightarrow \text{2y-4x=0}\]

\[\Rightarrow \text{y=2x}\]

$\therefore $ The equation of the line joining the given points is \[y=2x\].

ii. Find the equation of line joining \[\left( \text{3,1} \right)\] and \[\left( \text{9,3} \right)\] using determinants.

Ans: Let us assume a point, \[\text{P}\left( \text{x, y} \right)\] on the line joining points \[\text{A}\left( \text{3,1} \right)\] and \[\text{B}\left( \text{9,3} \right)\]. 

Then, the points \[\text{A}\],$B$ and $P$ are collinear.

Thus, the area of the triangle \[\text{ABP}\] will be zero.

\[\therefore \dfrac{\text{1}}{\text{2}}\left| \begin{matrix}   \text{3} & \text{1} & \text{1}  \\   \text{9} & \text{3} & \text{1}  \\   \text{x} & \text{y} & \text{1}  \\ \end{matrix} \right|\text{=0}\]

\[\Rightarrow \dfrac{\text{1}}{\text{2}}\left[ \text{3}\left( \text{3-y} \right)\text{-1}\left( \text{9-x} \right)\text{+1}\left( \text{9y-3x} \right) \right]\text{=0}\]

\[\Rightarrow \text{9-3y-9+x+9y-3x=0}\]

\[\Rightarrow \text{6y-2x=0}\]

\[\Rightarrow \text{x-3y=0}\]

$\therefore $ The equation of the line joining the given points is \[\text{x-3y=0}\] .

5. If the area of triangle is \[\text{35}\] square units with vertices \[\text{(2,-6)}\] , \[\left( \text{5,4} \right)\] and \[\left( \text{k,4} \right)\] . Then \[\text{k}\] is

1. \[\text{12}\]

2. \[\text{-2}\]

3. \[\text{-12,-2}\]

4. \[\text{12,-2}\]

Ans: Given vertices, \[\text{(2,-6)}\],\[\left( \text{5,4} \right)\] and \[\left( \text{k,4} \right)\]

The area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix}   \text{2} & \text{-6} & \text{1}  \\   \text{5} & \text{4} & \text{1}  \\   \text{k} & \text{4} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{2}\left( \text{4-4} \right)\text{+6}\left( \text{5-k} \right)\text{+1}\left( \text{20-4k} \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{50-10k} \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =25-5k}\]

Given, the area of the triangle is \[\text{35}\] square units.

Thus, we have:

\[\Rightarrow 25-5k=\pm 35\]

\[\Rightarrow 5\left( 5-k \right)=\pm 35\]

\[\Rightarrow 5-k=\pm 7\].

When \[5-k=7\]

\[\therefore k=-2\].

When \[5-k=-7\]

\[\therefore k=12\].

Hence, \[k=12,-2\] .

Thus, D. \[12,-2\] is the correct option.

Conclusion

The Class 12 Maths Chapter 4 Exercise 4.2 Solutions is pivotal for understanding the practical application of determinants in calculating the area of triangles. This exercise not only reinforces the theoretical knowledge of determinants but also highlights their significance in solving real-world geometric problems.

In previous years, questions from Class 12 Ex 4.2 have been prominent in exams, typically asking students to calculate the area of a triangle using given vertex coordinates. These questions often test the application of the determinant formula and the verification of results using determinant properties.

Class 12 Maths Chapter 4: Exercises Breakdown

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