NCERT Solutions for Class 12 Maths Chapter 4 - Exercise

Exercise- 4.3

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.

Introduction to Chapter 4 Determinants:

You must have learned in the earlier classes about matrix and matrices. For every square matrix, you can assign a real number. It is called the Determinant.

Determinant of Matrices of Order 1:

If A = [a ] be the matrix of order 1, then Determinant of A is said to be equal to a.

Determinant of Matrices of Order 2 ☓ 2:

For matrices with order 2, the Determinant is obtained through the multiplication of matrices.

Determinant of Matrices of Order 3 ☓ 3:

For matrices with order 3 ☓ 3, the Determinant is obtained by a series of steps explained in the NCERT solutions for Class 12 Maths Part 1 Chapter 4 Determinants. 

Properties of Determinants :

Chapter 4, Determinants, gives us a few properties.

• Property 1: If you interchange the rows and columns, the value of Determinants will not change. 

• Property 2: If two rows or 2 columns are interchanged, then the value of Determinants will change concerning sign.

• Property 3:  If two-row or columns are the same, then the value of Determinants is zero.

• Property 4: If each element in matrix is multiplied by a constant k, then the value of Determinants will also be multiplied by k. 

• Property 5: If the sum of two terms expresses elements, then the value of Determinants will also be expressed as the sum

Area of a Triangle :

As we have seen in earlier classes, the area of a triangle is half into base into height.

Here, the value of the area of the triangle is expressed as half of the value of the matrix of order 3 ☓ 3.

Exercise 4.3 has five questions based on these concepts.  

For detailed and correct solutions for Exercise 4.3, you can download our NCERT Solutions for Class 12 Maths Part 1 Chapter 4 Determinants. 

An Overview of Exercise 4.3 of Class 12 Maths NCERT Solutions

Area of Triangle in Determinant Form

When the coordinates of the triangle's vertices are given, the area of the triangle in determinant form is calculated in coordinate geometry. One of the most important applications of determinants is finding the area of a triangle in determinant form. The area of a triangle is usually calculated using the formula half the product of the triangle's base and altitude. However, if the triangle's height is unknown but its vertices are known, the determinant formula can be used to calculate the triangle's area.

The Formula for Finding the Area of a Triangle in Determinant Form

In determinant form, the formula for the area of a triangle gives a scalar value that can be positive or negative. However, because the area of a triangle can never be negative, we consider the determinant's absolute value as the triangle's area. If we have the vertices of a triangle ABC as A(x1, y1), B(x2, y2), and C(x3, y3) in a cartesian plane, then we can calculate the area of the triangle in determinant form with the below-given formula:

(image will be uploaded soon)

The Importance of Using NCERT Solutions for Class 12 Maths Part 1 Chapter 4- Determinants :

Class 12 is a fundamental level in the student's life. The future depends on these grades. Scoring better marks in Maths is a dream for many. Many students spend unnecessary time and effort just in cracking these answers. Your hurdle to success blows up here. With expert guidance! Vedantu solution describes all the questions asked in exercise 4.3 in a detailed manner as per CBSE guidelines. These NCERT Solutions for Maths part - 1 will help you get better marks in your Class 12. The curriculum is strictly taken from the NCERT Class 12 textbook. Our expert team of hardworking professionals is the reason behind developed and distinctive NCERT solutions.

All our answers are designed, and sample solutions are provided adhering to the NCERT textbook.

Our NCERT solutions give specific and to the point answers for the questions asked in NCERT Class 12 Maths Part 1 Chapter 4 Determinants exercise 4.3. 

Our experienced and highly qualified teachers have formulated these answers keeping in mind the situation of Class 12 students. With simple steps, exercise 4.3 is thoroughly solved. 

Benefits of Choosing Vedantu As Your Learning Mentor

Vedantu is responsible for carving the bright futures of many Class 12 students. The language of the NCERT solutions is kept simple and lucid for better understanding. All our solutions are in pdf versions. This makes it easy to download and accessible widely by the students. With step by step illustration of the solution for exercise 4.3, you are sure to score better marks in Maths Part 1 Class 12. For those students who would not want to miss any single chance of studying this year, we have devised our student-friendly app to view the solutions on the go.

With Vedantu, your success seems nearer than expected. Utilize your precious time this year, by going through our NCERT Solutions for Class 12 Maths Part 1 Chapter 4 Determinants Exercise 4.3. Our professionals strive hard in providing the best possible answers to the questions asked in the NCERT Class 12 Maths textbook. All our solutions are written exclusively for Class 12 students to understand and grasp, so don't feel the need for rote learning.

Chapter wise NCERT Solutions for Class 12 Maths

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

Access Other Exercises of Class 12 Maths Chapter 4

Why Vedantu?

Our well capable team of teachers formulates correct answers specific to the questions asked. PDF version of the solutions based on NCERT Class 12 Maths Part 1 Chapter 4. Vedantu app is developed for the better availability of answers to students of Class 12. Subscribe to our Web portal to gain access to numerous solutions. Our solutions are easy to understand the explanation of concepts that helps the students learn faster. You can also access our live classes that clear the doubts of all subjects in Class 12. We offer smarter learning methods for an intelligent generation.

Keeping all these points in mind, if you are struggling to score better marks in Maths Part 1 for Class 12, then Vedantu solutions are the right choice for you. Our NCERT solutions for Class 12 Maths Part 1 Chapter 4 Determinants are ideal for everyone who wants to pass Maths with flying colours. For a comfortable viewing experience, download the Vedantu app from the play store.