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

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. You are advised to solve more problems on your own to overcome the problems of silly mistakes. Also, check here for NCERT Solutions.

Also, see

• Determinants Exercise 4.1
• Determinants Exercise 4.2
• Determinants Exercise 4.4
• Determinants Exercise 4.5
• Determinants Exercise 4.6
• Determinants Miscellaneous Exercise

## Determinants Exercise:4.3

Question:1(i) Find area of the triangle with vertices at the point given in each of the following :

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

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

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

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

Question:1(ii) Find area of the triangle with vertices at the point given in each of the following :

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

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

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

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

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

Question:1(iii) Find area of the triangle with vertices at the point given in each of the following :

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

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 $|-15| = 15\ square\ units.$

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

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.

Question:3(i) Find values of k if area of triangle is 4 sq. units and vertices are

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

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.

Question:3(ii) Find values of k if area of triangle is 4 sq. units and vertices are

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

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 $k = 0$.

Question:4(i) Find equation of line joining $\small (1,2)$ and $\small (3,6)$ using determinants.

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 $\Rightarrow y=2x$.

Question:4(ii) Find equation of line joining $\small (3,1)$ and $\small (9,3)$ using determinants.

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 $x-3y = 0$.

Question:5 If the area of triangle is 35 sq units with vertices $\small (2,-6),(5,4)$ and $\small (k,4)$. Then k is

(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.

Also see-

• NCERT Solutions for Class 12 Maths Chapter 4

• NCERT Exemplar Solutions Class 12 Maths Chapter 4

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