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Q3(ii):
Find values of $k$ if area of triangle is 4 sq. units and vertices are (ii) $(–2, 0), (0, 4), (0, k)$

Solution :

Step 1: Theoretical Foundation for Area of a Triangle

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is defined by the determinant formula [Per the Coordinate Geometry definition of area]:

$\text{Area} = \frac{1}{2} \left| \det \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix} \right|$

Step 2: Substituting Given Coordinates

Given vertices: $(x_1, y_1) = (-2, 0)$, $(x_2, y_2) = (0, 4)$, $(x_3, y_3) = (0, k)$.

Given Area = $4$ sq. units. Therefore:

$4 = \frac{1}{2} \left| \det \begin{pmatrix} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{pmatrix} \right|$

$8 = \left| \det \begin{pmatrix} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{pmatrix} \right|$ [By multiplying both sides by 2]

Step 3: Evaluating the Determinant

Expanding the determinant along the first column (Column 1) for computational efficiency [By Laplace Expansion]:

$\det = -2 \begin{vmatrix} 4 & 1 \\ k & 1 \end{vmatrix} - 0 \begin{vmatrix} 0 & 1 \\ k & 1 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \\ 4 & 1 \end{vmatrix}$

$\det = -2(4(1) - k(1)) - 0 + 0$

$\det = -2(4 - k)$

$\det = -8 + 2k$

Step 4: Solving for $k$

Since the determinant is within an absolute value sign, we must account for both positive and negative scenarios [By the Definition of Absolute Value]:

$|2k - 8| = 8$

Case 1:

$2k - 8 = 8$

$2k = 16 \implies k = 8$

Case 2:

$2k - 8 = -8$

$2k = 0 \implies k = 0$

Visual Representation of Triangle Vertices

A(-2,0) B(0,4) C(0,k)

Final Solution: The values of $k$ are $0$ and $8$.


More Questions from Class 12 Mathematics Determinants EXERCISE 4.2


CBSE Solutions for Class 12 Mathematics Determinants


Chapters in CBSE - Class 12 Mathematics


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