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Q1(iii):
Find area of the triangle with vertices at the point given in each of the following: (iii) $(–2, –3), (3, 2), (–1, –8)$
Solution :
Given Vertices & Coordinate Setup
The vertices of the triangle are provided as follows:
- $A(x_1, y_1) = (-2, -3)$
- $B(x_2, y_2) = (3, 2)$
- $C(x_3, y_3) = (-1, -8)$
Step 1: Establishing the Determinant Formula
The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the determinant representation [Per the definition of area in coordinate geometry]:
$\text{Area} = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$
Step 2: Substituting Values into the Matrix
Substituting the given coordinates into the determinant structure:
$\text{Area} = \frac{1}{2} \begin{vmatrix} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \end{vmatrix}$
Step 3: Calculating the Determinant Value
Expanding the determinant along the first row ($R_1$):
$\Delta = -2 \begin{vmatrix} 2 & 1 \\ -8 & 1 \end{vmatrix} - (-3) \begin{vmatrix} 3 & 1 \\ -1 & 1 \end{vmatrix} + 1 \begin{vmatrix} 3 & 2 \\ -1 & -8 \end{vmatrix}$
Calculating the $2 \times 2$ determinants:
- $\begin{vmatrix} 2 & 1 \\ -8 & 1 \end{vmatrix} = (2)(1) - (-8)(1) = 2 + 8 = 10$
- $\begin{vmatrix} 3 & 1 \\ -1 & 1 \end{vmatrix} = (3)(1) - (-1)(1) = 3 + 1 = 4$
- $\begin{vmatrix} 3 & 2 \\ -1 & -8 \end{vmatrix} = (3)(-8) - (-1)(2) = -24 + 2 = -22$
Substituting back into the expansion:
$\text{Area} = \frac{1}{2} [ -2(10) + 3(4) + 1(-22) ]$
$\text{Area} = \frac{1}{2} [ -20 + 12 - 22 ]$
$\text{Area} = \frac{1}{2} [ -30 ]$
Step 4: Applying Absolute Magnitude
Since the area must be a positive scalar quantity [By definition of spatial area], we take the absolute value of the determinant result:
$\text{Area} = | -15 | = 15$
Visualization of Vertices
Final Solution: 15 \text{ square units}
More Questions from Class 12 Mathematics Determinants EXERCISE 4.2
- Q1(i): Find area of the triangle with vertices at the point given in each of the following: (i) $(1, 0), (6, 0), (4, 3)$
- Q1(ii): Find area of the triangle with vertices at the point given in each of the following: (ii) $(2, 7), (1, 1), (10, 8)$
- 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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