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Q1(i):
Find area of the triangle with vertices at the point given in each of the following: (i) $(1, 0), (6, 0), (4, 3)$
Solution :
Given Variables & Initial Setup
The vertices of the triangle are given as:
- $A(x_1, y_1) = (1, 0)$
- $B(x_2, y_2) = (6, 0)$
- $C(x_3, y_3) = (4, 3)$
Step 1: Establishing the Determinant Formula
The area $\Delta$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula: $$\Delta = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$$
Step 2: Substituting Values into the Determinant
Substituting the coordinates $(1, 0)$, $(6, 0)$, and $(4, 3)$ into the matrix expression: $$\Delta = \frac{1}{2} \begin{vmatrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix}$$
Step 3: Calculating the Determinant Value
We expand the determinant along the second column ($C_2$) to simplify the calculation, as it contains the most zeros: $$\Delta = \frac{1}{2} \left[ -(0) \begin{vmatrix} 6 & 1 \\ 4 & 1 \end{vmatrix} + (0) \begin{vmatrix} 1 & 1 \\ 4 & 1 \end{vmatrix} - (3) \begin{vmatrix} 1 & 1 \\ 6 & 1 \end{vmatrix} \right]$$ [Expanding along $C_2$ based on the cofactor expansion theorem]
Calculating the $2 \times 2$ determinant: $$\begin{vmatrix} 1 & 1 \\ 6 & 1 \end{vmatrix} = (1 \times 1) - (6 \times 1) = 1 - 6 = -5$$
Step 4: Final Scalar Multiplication
Substitute the evaluated determinant back into the area equation: $$\Delta = \frac{1}{2} \left[ -3 \times (-5) \right]$$ $$\Delta = \frac{1}{2} \times 15$$ $$\Delta = 7.5 \text{ square units}$$
Geometric Representation
Final Solution: The area of the triangle is 7.5 square units.
More Questions from Class 12 Mathematics Determinants EXERCISE 4.2
- Q1(ii): Find area of the triangle with vertices at the point given in each of the following: (ii) $(2, 7), (1, 1), (10, 8)$
- Q1(iii): Find area of the triangle with vertices at the point given in each of the following: (iii) $(–2, –3), (3, 2), (–1, –8)$
- Q2: Show that points A $(a, b + c)$, B $(b, c + a)$, C $(c, a + b)$ are collinear.
- Q3(i): Find values of $k$ if area of triangle is 4 sq. units and vertices are (i) $(k, 0), (4, 0), (0, 2)$
- Q3(ii): Find values of $k$ if area of triangle is 4 sq. units and vertices are (ii) $(–2, 0), (0, 4), (0, k)$
- Q4(i): (i) Find equation of line joining $(1, 2)$ and $(3, 6)$ using determinants.
- Q4(ii): (ii) Find equation of line joining $(3, 1)$ and $(9, 3)$ using determinants.
- Q5: If area of triangle is 35 sq units with vertices $(2, – 6), (5, 4)$ and $(k, 4)$. Then $k$ is
CBSE Solutions for Class 12 Mathematics Determinants
Chapters in CBSE - Class 12 Mathematics
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