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Q3(i):
Find values of $k$ if area of triangle is 4 sq. units and vertices are (i) $(k, 0), (4, 0), (0, 2)$
Solution :
Given Variables & Initial Setup
The vertices of the triangle are given as $A(k, 0)$, $B(4, 0)$, and $C(0, 2)$. The area of the triangle is given as $\Delta = 4$ sq. units.
The formula for the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ using determinants is:
$\Delta = \frac{1}{2} \left| \det \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix} \right|$
[Per the definition of Area of a triangle in Coordinate Geometry]
Step 1: Setting up the Determinant Equation
Substituting the given coordinates into the determinant expression:
$4 = \frac{1}{2} \left| \det \begin{pmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{pmatrix} \right|$
Multiplying both sides by 2, we obtain:
$8 = \left| \det \begin{pmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{pmatrix} \right|$
Step 2: Evaluating the Determinant
Expanding the determinant along the first column ($C_1$):
$\det = k(0(1) - 1(2)) - 4(0(1) - 1(2)) + 0(0(1) - 0(1))$
$\det = k(-2) - 4(-2) + 0$
$\det = -2k + 8$
[Per the Laplace Expansion Theorem for determinants]
Step 3: Solving for $k$
Since the area is an absolute value, we have:
$\left| -2k + 8 \right| = 8$
This leads to two distinct linear equations:
- $-2k + 8 = 8$
- $-2k + 8 = -8$
Case 1:
$-2k = 8 - 8$
$-2k = 0 \implies k = 0$
Case 2:
$-2k = -8 - 8$
$-2k = -16 \implies k = 8$
Visual Representation
Final Solution: The values of $k$ are $0$ and $8$.
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) $(2, 7), (1, 1), (10, 8)$
- 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(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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