NCERT Class 12 Mathematics — Determinants, Exercise 4.2. All 5 questions solved.

Exercise 4.2 gives determinants their first real job. The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is

$$\Delta = \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$$

Two consequences generate the whole exercise. If the three points are collinear there is no triangle, so the determinant is zero — that is the collinearity test in question 2. And running it backwards, forcing the determinant to vanish for a general point $(x, y)$ gives the equation of the line through two fixed points, which is question 4.

Key insight. The determinant is signed; an area is not. So when you are asked to find an area, take the absolute value at the end. When the area is given and you are solving for an unknown, set the determinant equal to $\pm$ the area — dropping the negative branch loses half the answers, which is exactly what questions 3 and 5 are checking.

Question 1

Find area of the triangle with vertices at the point given in each of the following.

Question 1 (i)

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

Solution. Substitute into the formula and expand along the first row.

$$\Delta = \frac{1}{2}\begin{vmatrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix} = \frac{1}{2}\Big[1(0 – 3) – 0(6 – 4) + 1(18 – 0)\Big]$$

$$= \frac{1}{2}(-3 + 18) = \frac{15}{2}$$

Question 1 (ii)

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

Solution.

$$\Delta = \frac{1}{2}\begin{vmatrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{vmatrix} = \frac{1}{2}\Big[2(1 – 8) – 7(1 – 10) + 1(8 – 10)\Big]$$

$$= \frac{1}{2}\big[-14 + 63 – 2\big] = \frac{47}{2}$$

Question 1 (iii)

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

Solution.

$$\Delta = \frac{1}{2}\begin{vmatrix} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \end{vmatrix} = \frac{1}{2}\Big[-2(2 + 8) + 3(3 + 1) + 1(-24 + 2)\Big]$$

$$= \frac{1}{2}\big[-20 + 12 – 22\big] = \frac{-30}{2} = -15$$

The determinant is negative, which only reflects the order in which the vertices were listed. Area is a positive quantity, so take the modulus.

Question 2

Show that points $\mathrm{A}(a,\, b+c)$, $\mathrm{B}(b,\, c+a)$, $\mathrm{C}(c,\, a+b)$ are collinear.

Solution. Three points are collinear exactly when the triangle they form has zero area, so it is enough to show the determinant vanishes.

$$\Delta = \frac{1}{2}\begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{vmatrix}$$

Expanding this directly is messy. Instead use a column operation, which does not change the value of a determinant: replace $\mathrm{C}_2$ by $\mathrm{C}_2 + \mathrm{C}_1$.

$$\Delta = \frac{1}{2}\begin{vmatrix} a & a+b+c & 1 \\ b & a+b+c & 1 \\ c & a+b+c & 1 \end{vmatrix}$$

Now the second column is $(a+b+c)$ times the third column. A determinant with two proportional columns is zero, so

$$\Delta = 0$$

and the three points are collinear.

Question 3

Find values of $k$ if area of triangle is 4 sq. units and vertices are:

Question 3 (i)

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

Solution. Set up the determinant with $k$ in place.

$$\Delta = \frac{1}{2}\begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = \frac{1}{2}\Big[k(0 – 2) – 0(4 – 0) + 1(8 – 0)\Big] = \frac{1}{2}(8 – 2k)$$

The area is given as 4, so the determinant may be $+4$ or $-4$:

$$\frac{1}{2}(8 – 2k) = \pm 4 \quad\Rightarrow\quad 8 – 2k = \pm 8$$

Taking each branch: $8 – 2k = 8$ gives $k = 0$, and $8 – 2k = -8$ gives $k = 8$.

Question 3 (ii)

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

Solution.

$$\Delta = \frac{1}{2}\begin{vmatrix} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{vmatrix} = \frac{1}{2}\Big[-2(4 – k) – 0(0 – 0) + 1(0 – 0)\Big] = k – 4$$

$$|k – 4| = 4 \quad\Rightarrow\quad k – 4 = \pm 4$$

Question 4

Find equation of line joining the given points using determinants.

Question 4 (i)

$(1, 2)$ and $(3, 6)$

Solution. Let $(x, y)$ be any point on the line. Then $(x, y)$, $(1, 2)$ and $(3, 6)$ are collinear, so the area of the triangle they form is zero — and that single equation is the line.

$$\frac{1}{2}\begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix} = 0$$

$$x(2 – 6) – y(1 – 3) + 1(6 – 6) = 0$$

$$-4x + 2y = 0 \quad\Rightarrow\quad y = 2x$$

Question 4 (ii)

$(3, 1)$ and $(9, 3)$

Solution. The same construction.

$$\begin{vmatrix} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{vmatrix} = 0$$

$$x(1 – 3) – y(3 – 9) + 1(9 – 9) = 0$$

$$-2x + 6y = 0 \quad\Rightarrow\quad x – 3y = 0$$

Question 5

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

Solution.

$$\Delta = \frac{1}{2}\begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} = \frac{1}{2}\Big[2(4 – 4) + 6(5 – k) + 1(20 – 4k)\Big]$$

$$= \frac{1}{2}\big[0 + 30 – 6k + 20 – 4k\big] = \frac{1}{2}(50 – 10k) = 25 – 5k$$

Setting this to $\pm 35$:

$$25 – 5k = 35 \Rightarrow k = -2, \qquad 25 – 5k = -35 \Rightarrow k = 12$$

Both are genuine, which rules out (A) and (B) — each gives only one root — and (C), which has the sign of the first root wrong.

Common mistakes

• Forgetting the $\tfrac{1}{2}$. The determinant is twice the area. It is easy to lose while concentrating on the $3\times3$ expansion.
• Dropping the negative branch when the area is given. Questions 3 and 5 each have two answers because the determinant may be $\pm$ twice the area. Question 5 offers the single-root answers as options (A) and (B) precisely to catch this.
• Reporting a negative area. In question 1(iii) the determinant is $-30$; the area is $15$. Take the modulus when finding an area — but not when solving for an unknown, where both signs matter.
• Expanding question 2 by brute force. Adding $\mathrm{C}_1$ to $\mathrm{C}_2$ makes two columns proportional and the answer immediate. A column operation of the form $\mathrm{C}_i \to \mathrm{C}_i + \mathrm{C}_j$ never changes a determinant’s value.
• Treating the line in question 4 as a separate formula to memorise. It is the same area determinant with a general point $(x, y)$ as one vertex, set to zero.

Practise next

• Exercise 4.3 — minors and cofactors, the bookkeeping behind the expansions used here.
• Exercise 4.1 — worth revisiting if the $3\times3$ expansions in question 1 felt slow.