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NCERT Solutions · Class 12 · Chapter 4: Determinants

NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2

Exercise 4.2: Area of a triangle. Area \(= \dfrac12\left|\begin{vmatrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{vmatrix}\right|\). Zero area means the points are collinear, which also gives the equation of the line through two points. When an area is given, use both signs of the determinant.

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Exercise 4.2 questions and solutions

Exercise 4.2, Question 1

Find the area of the triangle with vertices:
(i) \((1, 0), (6, 0), (4, 3)\)
Show solution
  1. \[\tfrac12[1(0 - 3) - 0 + 1(18 - 0)] = \tfrac12(15)\]
Answer: \(\tfrac{15}{2}\) square units
(ii) \((2, 7), (1, 1), (10, 8)\)
Show solution
  1. \[\tfrac12[2(1 - 8) - 7(1 - 10) + 1(8 - 10)] = \tfrac12(-14 + 63 - 2)\]
Answer: \(\tfrac{47}{2}\) square units
(iii) \((-2, -3), (3, 2), (-1, -8)\)
Show solution
  1. \[\begin{aligned}\tfrac12[-2(2 + 8) + 3(3 + 1) + 1(-24 + 2)] &= \tfrac12(-20 + 12 - 22) \\ &= -15\end{aligned}\]; take the absolute value.
Answer: 15 square units

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Exercise 4.2, Question 2

Show that \(A(a, b + c),\ B(b, c + a),\ C(c, a + b)\) are collinear.
Show solution
  1. \[\Delta = \dfrac12\begin{vmatrix}a & b + c & 1 \\ b & c + a & 1 \\ c & a + b & 1\end{vmatrix}\]
  2. Expand along column 3 (all 1s): \[\begin{vmatrix}b & c + a \\ c & a + b\end{vmatrix} - \begin{vmatrix}a & b + c \\ c & a + b\end{vmatrix} + \begin{vmatrix}a & b + c \\ b & c + a\end{vmatrix}\]
  3. \[= (ab + b^2 - c^2 - ca) - (a^2 + ab - bc - c^2) + (ac + a^2 - b^2 - bc) = 0\] (every term cancels), so \(\Delta = 0\).
Answer: Area 0, so the points are collinear

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Exercise 4.2, Question 3

Find k if the area is 4 square units and the vertices are:
(i) \((k, 0), (4, 0), (0, 2)\)
Show solution
  1. \[\begin{aligned}\tfrac12[k(0 - 2) - 0 + 1(8 - 0)] &= \tfrac12(8 - 2k) \\ &= 4 - k\end{aligned}\]
  2. \(4 - k = \pm4\).
Answer: \(k = 0\) or \(k = 8\)
(ii) \((-2, 0), (0, 4), (0, k)\)
Show solution
  1. \(\tfrac12[-2(4 - k) - 0 + 0] = k - 4\).
  2. \(k - 4 = \pm4\).
Answer: \(k = 0\) or \(k = 8\)

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Exercise 4.2, Question 4

Using determinants, find the equation of the line joining:
(i) \((1, 2)\) and \((3, 6)\)
Show solution
  1. A point \(P(x, y)\) is on the line iff the three points have zero area: \[\begin{vmatrix}x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1\end{vmatrix} = 0\]
  2. \[\begin{aligned}x(2 - 6) - y(1 - 3) + (6 - 6) &= -4x + 2y \\ &= 0\end{aligned}\]
Answer: \(y = 2x\)
(ii) \((3, 1)\) and \((9, 3)\)
Show solution
  1. \[\begin{aligned}\begin{vmatrix}x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1\end{vmatrix} &= x(1 - 3) - y(3 - 9) + (9 - 9) \\ &= -2x + 6y \\ &= 0\end{aligned}\]
Answer: \(x - 3y = 0\)

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Exercise 4.2, Question 5

Area 35 square units, vertices \((2, -6), (5, 4), (k, 4)\). Then k is: (A) 12 (B) \(-2\) (C) \(-12, -2\) (D) \(12, -2\)
Show solution
  1. \[\begin{aligned}\tfrac12[2(4 - 4) + 6(5 - k) + (20 - 4k)] &= \tfrac12(50 - 10k) \\ &= 25 - 5k\end{aligned}\]
  2. \(25 - 5k = \pm35 \Rightarrow k = -2\) or \(k = 12\).
Answer: (D)

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Determinants has 40 original board-style questions (MCQ, assertion–reason, short and long answers, case studies) with step mark schemes, revision notes and a four-step Route to 95. Three sample questions are open to everyone; a free account opens the rest.

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Determinants in our sample papers: Sample paper 1 (questions 3, 4, 7, 32) · Sample paper 2 (questions 4, 6, 7, 35) · Sample paper 3 (questions 4, 7, 32) · Sample paper 4 (questions 4, 5, 7, 35) · Sample paper 5 (questions 4, 6, 7, 32).

Also useful: free MCQs and case studies for Determinants · formulas for this chapter (Class 12 formula sheet, free PDF) · official CBSE board and sample papers · our sample papers with marking scheme · the Route to 95 plan

Textbook: NCERT Mathematics Class 12, Parts I and II (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in. Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log.