NCERT Solutions for Exercise 4.3 Class 12 Maths Chapter 4 - Determinants

# NCERT Solutions for Exercise 4.3 Class 12 Maths Chapter 4 - Determinants

Edited By Ramraj Saini | Updated on Dec 03, 2023 04:03 PM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Maths Chapter 4 Exercise 4.3

NCERT Solutions for Exercise 4.3 Class 12 Maths Chapter 4 Determinants are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for Class 12 Maths chapter 4 exercise 4.3 consists of questions related to finding the area of a triangle using determinants. In earlier classes, you have learned about finding areas of the triangle of given vertices. Now you can find that using determinants also. All the questions in exercise 4.3 Class 12 Maths are related to finding the area of the triangle using determinants only.

Since the area is a positive quantity, you should always take the absolute value of the determinant. Class 12th Maths chapter 4 exercise 4.3 exercise is very simple as well as important for board exams. As it includes a lot of calculations, the chances of silly mistakes are more here. 12th class Maths exercise 4.3 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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## Determinants Exercise: 4.3

$(1,0), (6,0), (4,3)$

We can find the area of the triangle with vertices $(1,0), (6,0), (4,3)$ by the following determinant relation:

$\triangle =\frac{1}{2} \begin{vmatrix} 1& 0 &1 \\ 6 & 0 &1 \\ 4& 3& 1 \end{vmatrix}$

Expanding using second column

$=\frac{1}{2} (-3) \begin{vmatrix} 1 &1 & \\ 6& 1 & \end{vmatrix}$

$= \frac{15}{2}\ square\ units.$

$(2,7), (1,1), (10,8)$

We can find the area of the triangle with given coordinates by the following method:

$\triangle = \begin{vmatrix} 2 &7 &1 \\ 1 & 1& 1\\ 10& 8 &1 \end{vmatrix}$

$=\frac{1}{2} \begin{vmatrix} 2 &7 &1 \\ 1 & 1& 1\\ 10& 8 &1 \end{vmatrix} = \frac{1}{2}\left [ 2(1-8)-7(1-10)+1(8-10) \right ]$

$= \frac{1}{2}\left [ 2(-7)-7(-9)+1(-2) \right ] = \frac{1}{2}\left [ -14+63-2 \right ] = \frac{47}{2}\ square\ units.$

$(-2,-3), (3,2), (-1,-8)$

Area of the triangle by the determinant method:

$Area\ \triangle = \frac{1}{2} \begin{vmatrix} -2 &-3 &1 \\ 3& 2 & 1\\ -1& -8 & 1 \end{vmatrix}$

$=\frac{1}{2}\left [ -2(2+8)+3(3+1)+1(-24+2) \right ]$

$=\frac{1}{2}\left [ -20+12-22 \right ] = \frac{1}{2}[-30]= -15$

Hence the area is equal to $|-15| = 15\ square\ units.$

If the area formed by the points is equal to zero then we can say that the points are collinear.

So, we have an area of a triangle given by,

$\triangle = \frac{1}{2} \begin{vmatrix} a &b+c &1 \\ b& c+a &1 \\ c& a+b & 1 \end{vmatrix}$

calculating the area:

$= \frac{1}{2}\left [ a\begin{vmatrix} c+a &1 \\ a+b& 1 \end{vmatrix} - (b+c)\begin{vmatrix} b & 1\\ c&1 \end{vmatrix}+1\begin{vmatrix} b &c+a \\ c&a+b \end{vmatrix} \right ]$

$= \frac{1}{2}\left [ a(c+a-a-b) - (b+c)(b-c)+1(b(a+b)-c(c+a)) \right ]$

$= \frac{1}{2}\left [ ac-ab - b^2+c^2+ab+b^2-c^2-ac \right ] = \frac{1}{2} \left [ 0 \right] = 0$

Hence the area of the triangle formed by the points is equal to zero.

Therefore given points $A (a, b+c), B (b,c+a), C (c,a+b)$ are collinear.

$(k,0), (4,0), (0,2)$

We can easily calculate the area by the formula :

$\triangle = \frac{1}{2} \begin{vmatrix} k &0 &1 \\ 4& 0& 1\\ 0 &2 & 1 \end{vmatrix} = 4\ sq.\ units$

$= \frac{1}{2}\left [ k\begin{vmatrix} 0 &1 \\ 2& 1 \end{vmatrix} -0\begin{vmatrix} 4 &1 \\ 0 & 1 \end{vmatrix}+1\begin{vmatrix} 4 &0 \\ 0& 2 \end{vmatrix} \right ]= 4\ sq.\ units$

$=\frac{1}{2}\left [ k(0-2)-0+1(8-0) \right ] = \frac{1}{2}\left [ -2k+8 \right ] = 4\ sq.\ units$

$\left [ -2k+8 \right ] = 8\ sq.\ units$ or $-2k +8 = \pm 8\ sq.\ units$

or $k = 0$ or $k = 8$

Hence two values are possible for k.

$(-2,0), (0,4), (0,k)$

The area of the triangle is given by the formula:

$\triangle = \frac{1}{2} \begin{vmatrix} -2 &0 &1 \\ 0 & 4 & 1\\ 0& k & 1 \end{vmatrix} = 4\ sq.\ units.$

Now, calculating the area:

$= \frac{1}{2} \left | -2(4-k)-0(0-0)+1(0-0) \right | = \frac{1}{2} \left | -8+2k \right | = 4$

or $-8+2k =\pm 8$

Therefore we have two possible values of 'k' i.e., $k = 8$ or $k = 0$.

As we know the line joining $\small (1,2)$ ,$\small (3,6)$ and let say a point on line $A\left ( x,y \right )$ will be collinear.

Therefore area formed by them will be equal to zero.

$\triangle = \frac{1}{2}\begin{vmatrix} 1 &2 &1 \\ 3& 6 &1 \\ x & y &1 \end{vmatrix} = 0$

So, we have:

$=1(6-y)-2(3-x)+1(3y-6x) = 0$

or $6-y-6+2x+3y-6x = 0 \Rightarrow 2y-4x=0$

Hence, we have the equation of line $\Rightarrow y=2x$.

We can find the equation of the line by considering any arbitrary point $A(x,y)$ on line.

So, we have three points which are collinear and therefore area surrounded by them will be equal to zero.

$\triangle = \frac{1}{2}\begin{vmatrix} 3 &1 &1 \\ 9& 3 & 1\\ x& y &1 \end{vmatrix} = 0$

Calculating the determinant:

$=\frac{1}{2}\left [ 3\begin{vmatrix} 3 &1 \\ y& 1 \end{vmatrix}-1\begin{vmatrix} 9 &1 \\ x& 1 \end{vmatrix}+1\begin{vmatrix} 9 &3 \\ x &y \end{vmatrix} \right ]$

$=\frac{1}{2}\left [ 3(3-y)-1(9-x)+1(9y-3x) \right ] = 0$

$\frac{1}{2}\left [ 9-3y-9+x+9y-3x \right ] = \frac{1}{2}[6y-2x] = 0$

Hence we have the line equation:

$3y= x$ or $x-3y = 0$.

(A) $\small 12$ (B) $\small -2$ (C) $\small -12,-2$ (D) $\small 12,-2$

Area of triangle is given by:

$\triangle = \frac{1}{2} \begin{vmatrix} 2 &-6 &1 \\ 5& 4 & 1\\ k& 4& 1 \end{vmatrix} = 35\ sq.\ units.$

or $\begin{vmatrix} 2 &-6 &1 \\ 5& 4 & 1\\ k& 4& 1 \end{vmatrix} = 70\ sq.\ units.$

$2\begin{vmatrix} 4 &1 \\ 4& 1 \end{vmatrix}-(-6)\begin{vmatrix} 5 &1 \\ k &1 \end{vmatrix}+1\begin{vmatrix} 5 &4 \\ k&4 \end{vmatrix} = 70$

$2(4-4) +6(5-k)+(20-4k) = \pm70$

$50-10k = \pm70$

$k = 12$ or $k = -2$

Hence the possible values of k are 12 and -2.

Therefore option (D) is correct.

## More About NCERT Solutions for Class 12 Maths chapter 4 exercise 4.3:-

In exercise 4.3 Class 12 Maths, there are 4 long answer types questions and one multiple-choice type question related to finding the area of the triangle. There are two solved examples given before Class 12 Maths ch 4 ex 4.3 that you can solve to get conceptual clarity. You should solve given exercise problems on your own. If you are facing difficulties while solving them, you can take help from NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.3.

Also Read| Determinants Class 12 Chapter 4 Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.3:-

• Class 12 Maths chapter 4 exercise 4.3 solutions are very helpful for the students who are facing difficulties while solving NCERT problems.
• NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.3 are designed following the guideline given by CBSE.
• You don't need to buy any additional books for board exams, NCERT textbook is enough for the CBSE board exam.
• You need to be thorough with NCERT textbook problems including solved examples.
• You can use these Class 12 Maths chapter 4 exercise 4.3 solutions for reference.
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## Key Features Of NCERT Solutions for Exercise 4.3 Class 12 Maths Chapter 4

• Comprehensive Coverage: The solutions encompass all the topics covered in ex 4.3 class 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 maths ex 4.3, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 ex 4.3 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this 12th class maths exercise 4.3 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for ex 4.3 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for class 12 maths ex 4.3 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

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## NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. Can we find area of triangle using determinant ?

Yes, we can find the area of the triangle using determinants if the vertices of the triangle are given.

Yes, here you will get NCERT Solutions for free. NCERT solutions for Class 12 Maths, Physics, Chemistry and Biology are available chapter wise.

3. Does CBSE provides NCERT books online ?

4. Where can I get NCERT solutions for class 6 to 10 ?

Here you will get NCERT Solutions for Class 6 to 10. The solutions for Science and Mathematics are given chapter wise. Students can follow NCERT syllabus for a better score to get good marks in CBSE annual exams.

5. what is the triangle formed by three collinear points ?

Triangle formed by three collinear points is a straight line, hence the area of such a triangle is zero.

6. What are the uses of NCERT solutions ?

NCERT solutions are useful for the students to get conceptual clarity.

7. Is the NCERT solutions helpful for competitive exams ?

NCERT solutions are helpful for competitive exams like JEE Main, NEET, VITEEE, etc.

8. How many chapters are there in the Class 12 Maths ?

There are 13 chapters starting from relations and functions to probability. Chapters from algebra, calculus are present.

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

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hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

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 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

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 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

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