Determinants 4.2 NCERT Solutions focuses on one of the most interesting applications of determinants in coordinate geometry i.e. finding the area of a triangle when the coordinates of its three vertices are given.
It also builds a strong foundation for identifying collinearity of three points lying on a straight line by setting the determinant equal to zero. This is a key idea used in deriving equations of lines passing through two given points with a variable third point (x, y).
Key Concepts
- Area of Triangle Using Determinants: If the vertices of a triangle are given as (x_1, y_1), (x_2, y_2), (x_3, y_3) , then its area is given by \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| .
- Determinant Form of Area: The same area can be written as \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right| , which gives a systematic method to compute area using determinants.
- Absolute Value of Area: Since area is always a positive quantity, the modulus (absolute value) of the determinant is taken in the final answer.
- Sign Consideration: If the determinant value is negative, it is still valid; only the magnitude matters when calculating area.
- Use of Positive and Negative Values: When the area of a triangle is already given and a coordinate is unknown, the determinant equation is solved by considering both positive and negative values, i.e., \pm \text{Area} , since the modulus is removed during algebraic calculation.
- Collinearity Condition: If three points are collinear, then the area of the triangle formed by them is zero, i.e., \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0 .
- Use in Equation of Line: By taking a variable point (x, y) along with two fixed points, the zero determinant condition helps derive the equation of the straight line passing through them.
- General Strategy: First form the determinant using given points, then either evaluate for area (use modulus) or set it equal to zero for collinearity problems.
Now, let’s begin with Question 1 of Determinants 4.2.
Question 1: Determinants 4.2
1. Find area of the triangle with vertices at the point given in each of the following:
(i) (1, 0), (6, 0), (4, 3)
Solution
Area of triangle using determinant formula:
\text{Area} = \frac{1}{2} \left| \begin{vmatrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix} \right|
Expanding along 1st row:
= \frac{1}{2} |1(0-3) - 0(6-4) + 1(18-0)|
= \frac{1}{2} |-3 + 18| = \frac{1}{2} \times 15
Area = \frac{15}{2}
Hence, area = 7.5 square units.
(ii) (2, 7), (1, 1), (10, 8)
Solution
\text{Area} = \frac{1}{2} \left| \begin{vmatrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{vmatrix} \right|
Expanding along row 1, we get
Area = \frac{1}{2} |2(1-8) - 7(1-10) + 1(8-10)|
= \frac{1}{2} |-14 + 63 - 2|
= \frac{1}{2} \times 47
or Area = \frac{47}{2}
Hence, area = 23.5 square units.
(iii) (–2, –3), (3, 2), (–1, –8)
Solution
\text{Area} = \frac{1}{2} \left| \begin{vmatrix} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \end{vmatrix} \right|
Expanding along 1st row:
= \frac{1}{2} |-2(2+8) - (-3)(3+1) + 1(-24+2)|
= \frac{1}{2} |-20 + 12 - 22|
(Note that the value of determinant is negative here, but Area is a postive quantity, so consider the absolute (positive) value only)
= \frac{1}{2} |-30|
= 15
Hence, area = 15 square units.
You may already be following Maths Better for important concepts and exam-oriented questions, like integration tricks and solving linear equations using the Matrix Method etc. Similarly, this NCERT Solutions series for Class 12 Maths will help strengthen your concepts and improve problem-solving skills.
Now, let’s move on to the next question of Determinants 4.2.
Question 2: Determinants Ex 4.2
2. Show that points A (a, b + c), B (b, c + a), C (c, a + b) are collinear.
Solution
To show that the points are collinear, we must find the area of the triangle formed by them. If the area is zero, the points are collinear.
Area of triangle using determinant:
\text{Area} = \frac{1}{2} \left| \begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{vmatrix} \right|
Expanding along the first row:
= \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|
Now evaluating further:
= \frac{1}{2} \left| a[(c+a) - (a+b)] - (b+c)[b - c] + [(b(a+b) - c(c+a))] \right|
Simplifying each term:
= \frac{1}{2} \left| a(c - b) - (b+c)(b-c) + (ab + b^2 - c^2 - ac) \right|
Using the identity:
(b+c)(b-c) = b^2 - c^2
Substitute:
= \frac{1}{2} \left| a(c-b) - (b^2 - c^2) + ab + b^2 - c^2 - ac \right|
Now cancelling terms:
= \frac{1}{2} |0|
= 0
Hence, the area is zero, so the points are collinear.
Expanding determinants becomes much easier once you understand the correct row or column selection.
Question 3: Determinants 4.2
3. Find values of k if area of triangle is 4 sq. units and vertices are:
(i) (k, 0), (4, 0), (0, 2)
Solution
Area of triangle using determinant:
\text{Area} = \frac{1}{2} \left| \begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} \right|
Expanding:
= \frac{1}{2} |k(0-2) - 0(4-0) + 1(8-0)|
= \frac{1}{2} |-2k + 8|
Given area = 4:
\frac{1}{2} |-2k + 8| = 4
|-2k + 8| = 8
Case 1: (consider positive)
-2k + 8 = 8 \Rightarrow k = 0
Case 2: (consider negative)
-2k + 8 = -8 \Rightarrow k = 8
Hence, k = 0 or 8.
(ii) (–2, 0), (0, 4), (0, k)
Solution
\text{Area} = \frac{1}{2} \left| \begin{vmatrix} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{vmatrix} \right|
Expanding:
= \frac{1}{2} |-2(4 - k) - 0(0 - 0) + 1(0 - 0)|
= \frac{1}{2} |-8 + 2k|
Given area = 4:
\frac{1}{2} |-8 + 2k| = 4
|-8 + 2k| = 8
Case 1: (taking positive value)
-8 + 2k = 8 \Rightarrow k = 8
Case 2: (taking negative value)
-8 + 2k = -8 \Rightarrow k = 0
Hence, k = 0 or 8.
Question 4: Ex 4.1 Determinants
4. (i) Find equation of line joining (1, 2) and (3, 6) using determinants.
Solution
For equation of line using determinant method, we use:
\begin{vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{vmatrix} = 0
Substituting points (1, 2) and (3, 6):
\begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix} = 0
Expanding along first row:
x(2-6) - y(1-3) + 1(6-6) = 0
or -4x + 2y = 0
or 2y = 4x
i.e. y = 2x
Hence, equation of line is y = 2x.
(ii) Find equation of line joining (3, 1) and (9, 3).
Solution
Using determinant method:
\begin{vmatrix} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{vmatrix} = 0
Expanding:
x(1-3) - y(3-9) + 1(9-9) = 0
-2x + 6y + 0 = 0
Dividing by 2:
-x + 3y = 0
Hence, equation of line is x – 3y = 0.
NCERT Class 12 Maths has a total of 61 exercises across Part 1 and Part 2 including the miscellaneous ones. I’ll cover each exercise one by one with clear explanations and exam-oriented solutions. Many of these questions are also available in video format on my YouTube channel, @Mathsbetter.
Try solving the next question on your own first, and then compare your approach with the solution given below.
Question 5: Determinants 4.2
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
Solution
Area of triangle using determinant formula:
\text{Area} = \frac{1}{2} \left| \begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} \right|
Expanding along first row:
= \frac{1}{2} \left| 2(4-4) - (-6)(5-k) + 1(20-4k) \right|
Simplifying:
= \frac{1}{2} |0 + 6(5-k) + 20 - 4k|
= \frac{1}{2} |30 - 6k + 20 - 4k|
So, Area = \frac{1}{2} |50 - 10k|
But given area = 35
Therefore,
\frac{1}{2} |50 - 10k| = 35
or |50 - 10k| = 70
or 10|5 - k| = 70
i.e. |5 - k| = 7
Case 1: (positive)
5 - k = 7 \Rightarrow k = -2
Case 2: (negative)
5 - k = -7 \Rightarrow k = 12
Hence, k = 12 or –2.
Correct option: (D) 12, –2
Common Mistakes to Avoid
- Forgetting absolute value in area: While finding the area of a triangle using determinants, students often forget that the final answer must be taken in modulus form \left| \cdot \right| , since area is always positive.
- Not using ± when area is given: When the area is provided and coordinates are unknown, students incorrectly use only the positive value of the determinant instead of considering both cases \pm \text{Area} .
- Incorrect collinearity assumption: Some students assume points are collinear without verifying that the determinant equals zero, which is the correct condition.
- Sign mistakes in determinant expansion: Errors commonly occur while applying alternating signs during expansion of 3×3 determinants, leading to wrong area or equation results.
- Incorrect substitution of coordinates: Careless substitution of point coordinates into the determinant form leads to incorrect expressions and final answers.
- Mixing area and line equations: Students sometimes confuse the area formula with the line equation determinant form \begin{vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{vmatrix} = 0 , especially in collinearity-based questions.
Continue Learning
- Strengthen understanding of determinant-based area formula and its geometric interpretation in coordinate geometry.
- Build fluency in identifying collinearity conditions using determinant equal to zero.
- Learn how determinant method is used to derive equations of lines from two given points.
- Develop speed and accuracy in expanding 3×3 determinants as it is frequently used in higher topics.
Explore More
You may also explore more NCERT Solutions, important questions and concept-based explanations on Maths Better to strengthen your preparation step-by-step.
All the best and keep learning 👍
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