# NCERT Solutions Class 12 Maths Chapter 4 Exercise 4.3

Class 12 plays a very important role in the educational career of a student. The board examinations in Class 12 form the basis for admission into several colleges and universities. The marks obtained in the board examinations are an important criterion for securing admissions to various courses. Hence, students need to score good marks in the board examination. However, preparing for the board examination is not an easy task. It requires a lot of hard work and smart preparation. Especially, for a subject like Mathematics, students have to practice a lot of questions. That is why it is recommended by teachers to practice NCERT questions. The NCERT textbook is also recommended by the CBSE for its board examinations. Hence, students should solve the NCERT exercises.

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## NCERT Solutions Class 12 Maths Chapter 4 Exercise 4.3

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## Access NCERT Solutions for Class 12 Maths Chapter 4 – Determinants

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### Introduction to Chapter 4 Determinants:

A determinant is a numerical value that is a function of the elements of a square matrix. The chapter on determinants is a high-scoring and crucial chapter from the point of view of the board examinations. Illustrations help students to understand the different sub-topics covered in the chapter. Students must practise the exercises given at the end of each chapter to analyse their understanding of the topic. Having access to the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.3 helps students self-evaluate their understanding, and identify their strengths and weaknesses.

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### Area of a Triangle :

Students can find the area of a triangle with the help of determinants. They should solve the questions related to finding the area of a triangle from the exercises to get a good grasp of the concept. Moreover, the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.3 can guide students in finding the area of a triangle for a clear and better understanding.

### An Overview of Exercise 4.3 of Class 12 Maths NCERT Solutions

Students learn to find the area of a triangle whose coordinates of vertices are given. From an examination point of view, it comes across as an essential exercise. The conceptual understanding and its application in this exercise play an important role. Access to the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.3 available on Extramarks, benefits students in multiple ways. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.3 assists students in solving difficult questions by providing easy and stepwise solutions to them. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.3 guides the students in developing an ability to solve questions with ease.

### Area of Triangle in Determinant Form

The area of a triangle in determinant form is an essential exercise that helps the students to find the area of a triangle. It aids students to understand the ways to find the area of a triangle using determinants. It remains an easy and high-scoring exercise for students. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.3 available at Extramarks are detailed and genuine solutions. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.3 also help the students in achieving good results in the examination. Also, the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.3 reinforce their fundamentals and will be an important part of their overall preparation.

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### Chapter 4 – Determinants of  Other Exercises

 Chapter 4 – Determinants of Other Exercises Exercise 4.1 8 Questions & Solutions (3 Short Answers, 5 Long Answers) Exercise 4.2 10 Questions & Solutions (4 Short Answers, 10 Long Answers) Exercise 4.4 5 Questions & Solutions (2 Short Answers, 3 Long Answers) Exercise 4.5 18 Questions & Solutions (4 Short Answers, 14 Long Answers) Exercise 4.6 16 Questions & Solutions (3 Short Answers, 13 Long Answers)

Q.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)

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Theare}\mathrm{a}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}\left(1,0\right),\left(6,0\right),\left(4,3\right)\\ \mathrm{\Delta }=\frac{1}{2}|\begin{array}{l}101\\ 601\\ 431\end{array}|\\ =\frac{1}{2}\left\{1\left(0-3\right)-0\left(6-4\right)+1\left(18-0\right)\right\}\\ =\frac{1}{2}\left(-3-0+18\right)\\ =\frac{1}{2}\left(15\right)\\ =\frac{15}{2}\mathrm{ }\mathrm{sq}.\mathrm{ }\mathrm{units}\\ \left(\mathrm{ii}\right)\mathrm{Theare}\mathrm{a}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}\left(2,7\right),\left(1,1\right),\left(10,8\right)\\ \mathrm{\Delta }=\frac{1}{2}|\begin{array}{l}271\\ 111\\ 1081\end{array}|\\ =\frac{1}{2}\left\{2\left(1-8\right)-7\left(1-10\right)+1\left(8-10\right)\right\}\\ =\frac{1}{2}\left(-14+63-2\right)\\ =\frac{1}{2}\left(47\right)\\ =\frac{47}{2}\mathrm{ }\mathrm{sq}.\mathrm{ }\mathrm{units}\\ \left(\mathrm{iii}\right)\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}\left(-2,-3\right),\\ \left(3,2\right),\left(-1,-8\right)\\ \mathrm{\Delta }=\frac{1}{2}|\begin{array}{l}-2-31\\ 3 \mathrm{ }21\\ -1-81\end{array}|\\ =\frac{1}{2}\left\{-2\left(2+8\right)-\left(-3\right)\left(3+1\right)+1\left(-24+2\right)\right\}\\ =\frac{1}{2}\left(-20+12-22\right)\\ =\frac{1}{2}\left(-30\right)\\ =-15\mathrm{ }\mathrm{sq}.\mathrm{ }\mathrm{units}\\ \mathrm{Since},\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{can}\mathrm{never}\mathrm{be}\mathrm{negative}.\\ \mathrm{So},\mathrm{area}\mathrm{of}\mathrm{triangle}= 15\mathrm{square}\mathrm{units}\mathrm{.}\end{array}$

Q.2 Show that points A(a, b + c), B(b, c + a), C(c, a + b) are collinear.

Ans

$\begin{array}{l}\mathrm{Theare}\mathrm{a}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}\mathrm{A}\left(\mathrm{a},\mathrm{b}+\mathrm{c}\right),\\ \mathrm{B}\left(\mathrm{b},\mathrm{c}+\mathrm{a}\right),\mathrm{C}\left(\mathrm{c},\mathrm{a}+\mathrm{b}\right)\\ \mathrm{\Delta }=\frac{1}{2}|\begin{array}{ccc}\mathrm{a}& \mathrm{b}+\mathrm{c}& 1\\ \mathrm{b}& \mathrm{c}+\mathrm{a}& 1\\ \mathrm{c}& \mathrm{a}+\mathrm{b}& 1\end{array}|\\ =\frac{1}{2}\left\{\mathrm{a}\left(\mathrm{c}+\mathrm{a}-\mathrm{a}-\mathrm{b}\right)-\left(\mathrm{b}+\mathrm{c}\right)\left(\mathrm{b}-\mathrm{c}\right)+1\left(\mathrm{ba}+{\mathrm{b}}^{2}-{\mathrm{c}}^{2}-\mathrm{ca}\right)\right\}\\ =\frac{1}{2}\left\{\mathrm{ac}-\mathrm{ab}-\left({\mathrm{b}}^{2}-{\mathrm{c}}^{2}\right)+\left(\mathrm{ba}+{\mathrm{b}}^{2}-{\mathrm{c}}^{2}-\mathrm{ca}\right)\right\}\\ =\frac{1}{2}\left(\overline{)\mathrm{ac}}-\overline{)\mathrm{ab}}-\overline{){\mathrm{b}}^{2}}+\overline{){\mathrm{c}}^{2}}+\overline{)\mathrm{ba}}+\overline{){\mathrm{b}}^{2}}-\overline{){\mathrm{c}}^{2}}-\overline{)\mathrm{ca}}\right)\\ =0\\ \mathrm{Since},\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{A},\mathrm{B}\mathrm{and}\mathrm{C}\mathrm{is}\mathrm{zero},\mathrm{so}\mathrm{these}\\ \mathrm{points}\mathrm{are}\mathrm{collinear}\mathrm{.}\end{array}$

Q.3 Find the value of k if area of triangle is 4 sq. units and vertices are

(i) (k, 0), (4, 0), (0, 2) (ii) (–2, 0), (0, 4), (0, k)

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{ }\mathrm{Given}:\mathrm{area}\mathrm{of}\mathrm{triangle}=±4 \mathrm{ }\left(\begin{array}{l}\because \mathrm{Since}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{is}\mathrm{absolute}\\ \mathrm{value}\mathrm{of}\mathrm{\Delta }\end{array}\right)\\ \mathrm{Then}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}\left(\mathrm{k},0\right),\\ \left(4,0\right),\left(0,2\right)\\ \mathrm{\Delta }=\frac{1}{2}|\begin{array}{l}\mathrm{k}01\\ 401\\ 021\end{array}|\\ ±4=\frac{1}{2}\left\{\mathrm{k}\left(0-2\right)-0\left(4-0\right)+1\left(8-0\right)\right\}\\ ±4=\frac{1}{2}\left(-2\mathrm{k}-0+8\right)\\ ±8=-2\mathrm{k}+8\\ ⇒\mathrm{ }-2\mathrm{k}+8=8 \mathrm{or} -2\mathrm{k}+8=-8\\ ⇒\mathrm{k}=0\mathrm{}\mathrm{or}\mathrm{k}=\frac{-16}{-2}=8\\ \mathrm{Thus},\mathrm{value}\mathrm{of}\mathrm{k}\mathrm{is}\mathrm{either}0\mathrm{or}8\mathrm{.}\\ \left(\mathrm{ii}\right) \mathrm{Given}:\mathrm{area}\mathrm{of}\mathrm{triangle}=±4 \mathrm{ }\left(\begin{array}{l}\because \mathrm{Since}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{is}\mathrm{absolute}\\ \mathrm{value}\mathrm{of}\mathrm{\Delta }\end{array}\right)\\ \mathrm{Then}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}\left(-2,0\right),\\ \left(0,4\right),\left(0,\mathrm{k}\right)\\ \mathrm{\Delta }=\frac{1}{2}|\begin{array}{l}-201\\ \mathrm{ }041\\ \mathrm{ }0\mathrm{k}1\end{array}|\\ ±4=\frac{1}{2}\left\{-2\left(4-\mathrm{k}\right)-0\left(0-0\right)+1\left(0-0\right)\right\}\\ ±4=\frac{1}{2}\left(-8+2\mathrm{k}-0+0\right)\\ ±8=2\mathrm{k}-8\\ ⇒\mathrm{ }2\mathrm{k}-8=8 \mathrm{or} 2\mathrm{k}-8=-8\\ ⇒\mathrm{k}=\frac{16}{2}=\mathrm{ }8\mathrm{}\mathrm{or}\mathrm{k}=\frac{0}{2}=0\\ \mathrm{Thus},\mathrm{value}\mathrm{of}\mathrm{k}\mathrm{is}\mathrm{either}8\mathrm{or}0\mathrm{.}\end{array}$

Q.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.

Ans

$\begin{array}{l}\left(i\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Point A}\left(1,2\right)\text{and B}\left(3,6\right)\text{are joining each other and a point P}\left(x,y\right)\text{lies}\\ \text{on it, so A, P and B are collinear and area of triangle formed by these}\\ \text{points is zero}\text{. Then}\\ Area\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\text{triangle ABP}=\text{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\begin{array}{l}121\\ 36\text{}1\\ xy\text{}1\end{array}|=0\\ ⇒\frac{1}{2}\left\{1\left(6-y\right)-2\left(3-x\right)+1\left(3y-6x\right)\right\}=0\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}6-y-6+2x+3y-6x=0\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2y-4x=0\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=2x\\ \text{Thus, the equation of line joining two points (1, 2) and}\left(3,6\right)\\ is\text{​}\text{y}=\text{2x}\text{.}\\ \left(ii\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Point A}\left(3,\text{\hspace{0.17em}}1\right)\text{and B}\left(9,3\right)\text{are joining each other and a point P}\left(x,y\right)\text{lies}\\ \text{on it, so A, P and B are collinear and area of triangle formed by these}\\ \text{points is zero}\text{. Then}\\ Area\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\text{triangle ABP}=\text{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\begin{array}{l}31\text{}1\\ 9\text{}31\\ x\text{}y1\end{array}|=0\\ ⇒\frac{1}{2}\left\{3\left(3-y\right)-1\left(9-x\right)+1\left(9y-3x\right)\right\}=0\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}9-3y-9+x+9y-3x=0\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}6y-2x=0\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\text{\hspace{0.17em}}y=x\\ \text{Thus, the equation of line joining two points (1, 2) and}\left(3,6\right)\\ is\text{​}\text{x}-\text{3y}=\text{0}\text{.}\end{array}$

Q.5 If area of triangle is 35 sq. units with vertices (2, -6), (5, 4) and (k, 4). Then k is
(A) 12 (B) –2 (C) –12, –2 (D) 12, –2

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{area}\mathrm{of}\mathrm{triangle}=± 35 \left(\begin{array}{l}\because \mathrm{Since}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{is}\mathrm{absolute}\\ \mathrm{value}\mathrm{of}\mathrm{\Delta }\end{array}\right)\\ \mathrm{Then}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}\left(2,-6\right),\\ \left(5,4\right),\left(\mathrm{k},4\right)\\ \mathrm{\Delta }=\frac{1}{2}|\begin{array}{l}2-61\\ 5 \mathrm{ }41\\ \mathrm{k} 41\end{array}|\\ ±35=\frac{1}{2}\left\{2\left(4-4\right)-\left(-6\right)\left(5-\mathrm{k}\right)+1\left(20-4\mathrm{k}\right)\right\}\\ ±35=\frac{1}{2}\left(0+30-6\mathrm{k}+20-4\mathrm{k}\right)\\ ±70=-10\mathrm{k}+50\\ ⇒\mathrm{ }-10\mathrm{k}+50=70 \mathrm{or} -10\mathrm{k}+50=-70\\ ⇒\mathrm{k}=\frac{20}{-10}=-2\mathrm{or}\mathrm{k}=\frac{-120}{-10}=12\\ \mathrm{Thus},\mathrm{value}\mathrm{of}\mathrm{k}\mathrm{is}\mathrm{either}12\mathrm{or}-2.\\ \mathrm{Thus},\mathrm{option}\left(\mathrm{D}\right)\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

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Students must practice the NCERT textbook daily to help finish the examination paper in the allotted period. Lack of practice remains one of the biggest reasons for students’ inability to solve the exam in the allotted time frame. Furthermore, the exam paper contains questions from the NCERT textbook. Having access to the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.3 helps students learn how to structure the answers to become efficient for timely completion of the question paper in the examination.

## 2. Does Extramarks provide NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.3 and other chapters for Class 12?

Yes, students can access the Class 12 Maths Chapter 4 Exercise 4.3 Solutions on the Extramarks’ website. The solutions are curated by academic experts, making them a reliable source of knowledge. The solutions are explained step-wise and are aligned with the answers mentioned in the NCERT textbooks to help students have an immersive experience.

## 3. Do the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.3 and other exercises aid students to absorb the concept of the different chapters?

For senior classes like Class 12, NCERT solutions are vital for the preparation for the board examinations. Learning modules like the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.3 allows students to have ample practice before appearing for any examination or test. Answers are not enough because students need to understand the steps to reach the answer. The clarity of the process and steps can ensure that the students develop the ability to solve any question that comes their way.