# NCERT Solutions Class 12 Maths Chapter 4 Exercise 4.3

## NCERT Solutions Class 12 Maths Chapter 4 Exercise 4.3

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

### 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:

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### Access Other Exercises of Class 12 Maths Chapter 4

 Chapter 4 Determinants  All Exercises in PDF Format Exercise 4.1 8 Questions and Solutions Exercise 4.2 16 Questions and Solutions Exercise 4.4 5 Questions and Solutions Exercise 4.5 18 Questions and Solutions Exercise 4.6 16 Questions and Solutions

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