NCERT Solutions Class 12 Maths Chapter 4 Ex 4.2 | LearnCBSEHub

Area & Collinearity

NCERT EXERCISE 4.2 • Q1 — Q5 COMPLETE

💡 Key Formula: Area of Triangle

[Image of area of triangle using determinants formula]

If vertices are $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$, then Area ($\Delta$) is given by:

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

Note: Since area is always positive, we take the absolute value. If points are collinear, Area = 0.

Question 01

Find area of the triangle with vertices: (i) (1, 0), (6, 0), (4, 3)

$$\text{Area} = \frac{1}{2} \begin{vmatrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix}$$ Expanding along $C_2$ (containing two zeros): $$\text{Area} = \frac{1}{2} \left[ -0 + 0 – 3 \begin{vmatrix} 1 & 1 \\ 6 & 1 \end{vmatrix} \right]$$ $$\text{Area} = \frac{1}{2} [ -3 (1 – 6) ] = \frac{1}{2} [ -3(-5) ] = \frac{15}{2}$$

Result: 7.5 sq. units

Question 02 • Collinearity

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

Points are collinear if the Area of the triangle formed by them is 0.

$$\Delta = \frac{1}{2} \begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{vmatrix}$$ Applying $C_2 \to C_2 + C_1$: $$\Delta = \frac{1}{2} \begin{vmatrix} a & a+b+c & 1 \\ b & a+b+c & 1 \\ c & a+b+c & 1 \end{vmatrix}$$ Taking $(a+b+c)$ common from $C_2$: $$\Delta = \frac{a+b+c}{2} \begin{vmatrix} a & 1 & 1 \\ b & 1 & 1 \\ c & 1 & 1 \end{vmatrix}$$

Since $C_2$ and $C_3$ are identical, the determinant value is 0.

✅ Hence, points are collinear.

Question 03

Find $k$ if area is 4 sq. units and vertices are (k, 0), (4, 0), (0, 2).

Given Area = 4. We use $\pm 4$ for calculation:

$$\frac{1}{2} \begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = \pm 4$$ Expanding along $C_2$: $$\frac{1}{2} [ -2 (k – 4) ] = \pm 4 \implies -(k – 4) = \pm 4$$ Case 1: $-k + 4 = 4 \implies k = 0$
Case 2: $-k + 4 = -4 \implies k = 8$

Result: $k = 0, 8$

Question 04 • Equation of Line

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

Let $P(x, y)$ be any point on the line. Then points $(x, y), (1, 2),$ and $(3, 6)$ are collinear ($\text{Area} = 0$).

$$\frac{1}{2} \begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix} = 0$$ $$x(2 – 6) – y(1 – 3) + 1(6 – 6) = 0$$ $$-4x + 2y = 0 \implies 2y = 4x$$

Equation: $y = 2x$

Question 05 • MCQ

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

$$\frac{1}{2} \begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} = \pm 35$$ $$2(4-4) – (-6)(5-k) + 1(20-4k) = \pm 70$$ $$30 – 6k + 20 – 4k = \pm 70 \implies 50 – 10k = \pm 70$$ Case 1: $50 – 10k = 70 \implies -10k = 20 \implies k = -2$
Case 2: $50 – 10k = -70 \implies -10k = -120 \implies k = 12$

Correct Option: (D) 12, -2