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Q1(ii):
Find area of the triangle with vertices at the point given in each of the following: (ii) $(2, 7), (1, 1), (10, 8)$
Solution :
Step 1: Theoretical Foundation
The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is determined by the absolute value of the determinant expression [Per the Area of Triangle Determinant Formula]:
$\text{Area} = \frac{1}{2} \left| \det \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix} \right|$
Step 2: Defining Coordinate Variables
Given the vertices $A(2, 7)$, $B(1, 1)$, and $C(10, 8)$, we map the variables as follows:
- $(x_1, y_1) = (2, 7)$
- $(x_2, y_2) = (1, 1)$
- $(x_3, y_3) = (10, 8)$
Step 3: Constructing the Determinant Matrix
Substitute the coordinates into the formula:
$\text{Area} = \frac{1}{2} \left| \det \begin{pmatrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{pmatrix} \right|$
Step 4: Algebraic Expansion along Row 1
Expand the determinant using the Laplace expansion method [By Row Expansion Rule]:
$\det = 2 \begin{vmatrix} 1 & 1 \\ 8 & 1 \end{vmatrix} - 7 \begin{vmatrix} 1 & 1 \\ 10 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & 1 \\ 10 & 8 \end{vmatrix}$
Calculate the $2 \times 2$ determinants:
- $\begin{vmatrix} 1 & 1 \\ 8 & 1 \end{vmatrix} = (1)(1) - (8)(1) = 1 - 8 = -7$
- $\begin{vmatrix} 1 & 1 \\ 10 & 1 \end{vmatrix} = (1)(1) - (10)(1) = 1 - 10 = -9$
- $\begin{vmatrix} 1 & 1 \\ 10 & 8 \end{vmatrix} = (1)(8) - (10)(1) = 8 - 10 = -2$
Step 5: Final Evaluation
Combine the results:
$\det = 2(-7) - 7(-9) + 1(-2)$
$\det = -14 + 63 - 2$
$\det = 47$
$\text{Area} = \frac{1}{2} |47| = 23.5 \text{ square units}$
Visual Representation
Final Solution: 23.5 \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(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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