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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 Komal Miglani | Updated on Apr 19, 2025 08:53 AM IST | #CBSE Class 12th

Imagine you're given three points in a plane and asked to find the area of the triangle they form. Instead of using traditional geometry formulas, determinants offer a powerful method to compute this area efficiently. Class 12 Maths Chapter 4 – Determinants, Exercise 4.2 of NCERT introduces the application of determinants in finding the area of a triangle when the coordinates of its vertices are known. This exercise enhances understanding of how algebraic concepts like determinants can be applied to solve geometric problems. This article on NCERT Solutions for Exercise 4.2 Class 12 Maths Chapter 4 - Determinants, offers clear and step-by-step solutions for the exercise problems.

Class 12 Maths Chapter 4 Exercise 4.2 Solutions: Download PDF

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Determinants Exercise: 4.2

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)

Answer:

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

△=12|101601431|

Expanding using second column

=12(−3)|1161|

=152 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)

Answer:

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

△=|2711111081|

=12|2711111081|=12[2(1−8)−7(1−10)+1(8−10)]

=12[2(−7)−7(−9)+1(−2)]=12[−14+63−2]=472 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)

Answer:

Area of the triangle by the determinant method:

Area △=12|−2−31321−1−81|

=12[−2(2+8)+3(3+1)+1(−24+2)]

=12[−20+12−22]=12[−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.

Answer:

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,

△=12|ab+c1bc+a1ca+b1|

calculating the area:

=12[a|c+a1a+b1|−(b+c)|b1c1|+1|bc+aca+b|]

=12[a(c+a−a−b)−(b+c)(b−c)+1(b(a+b)−c(c+a))]

=12[ac−ab−b2+c2+ab+b2−c2−ac]=12[0]=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)

Answer:

We can easily calculate the area by the formula :

△=12|k01401021|=4 sq. units

=12[k|0121|−0|4101|+1|4002|]=4 sq. units

=12[k(0−2)−0+1(8−0)]=12[−2k+8]=4 sq. units

[−2k+8]=8 sq. units or −2k+8=±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)

Answer:

The area of the triangle is given by the formula:

△=12|−2010410k1|=4 sq. units.

Now, calculating the area:

=12|−2(4−k)−0(0−0)+1(0−0)|=12|−8+2k|=4

or −8+2k=±8

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

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

Answer:

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

Therefore area formed by them will be equal to zero.

△=12|121361xy1|=0

So, we have:

=1(6−y)−2(3−x)+1(3y−6x)=0

or 6−y−6+2x+3y−6x=0⇒2y−4x=0

Hence, we have the equation of line ⇒y=2x.

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

Answer:

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.

△=12|311931xy1|=0

Calculating the determinant:

=12[3|31y1|−1|91x1|+1|93xy|]

=12[3(3−y)−1(9−x)+1(9y−3x)]=0

12[9−3y−9+x+9y−3x]=12[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 (2,−6),(5,4) and (k,4). Then k is

(A) 12 (B) −2 (C) −12,−2 (D) 12,−2

Answer:

Area of triangle is given by:

△=12|2−61541k41|=35 sq. units.

or |2−61541k41|=70 sq. units.

2|4141|−(−6)|51k1|+1|54k4|=70

2(4−4)+6(5−k)+(20−4k)=±70

50−10k=±70

k=12 or k=−2

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

Therefore option (D) is correct.


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Background wave

Topics Covered in Chapter 4, Determinants: Exercise 4.2

Here are the main topics covered in NCERT Class 12 Chapter 4, Determinants: Exercise 4.2:

1. Area of a Triangle Using Determinants:

Given the coordinates of the vertices of a triangle as A(x1,y1),B(x2,y2), and C(x3,y3), the area of the triangle can be calculated using the determinant:

Area =12|x1y11x2y21x3y31|

This formula utilizes the determinant of a 3×3 matrix formed by the coordinates of the triangle's vertices.

2. Collinearity of Points:

If the area calculated using the above determinant is zero, it implies that the three points are collinear, i.e., they lie on a straight line and do not form a triangle.

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Frequently Asked Questions (FAQs)

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.

2. Can I get NCERT solutions for free ?

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 ?

Yes, You can download the NCERT book from their official website.

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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Manasvini Gupta 30 January,2025
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Kanishka kaushikii 17 September,2024
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