(D) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula:$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3)

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(D) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula:$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$Given vertices are $A(1, 3)$,$B(0, 0)$,and $C(k, 0)$,and the area is $3$ sq. units.Substituting the values into the formula:$3 = \frac{1}{2} |1(0 - 0) + 0(0 - 3) + k(3 - 0)|$$3 = \frac{1}{2} |0 + 0 + 3k|$$3 = \frac{1}{2} |3k|$$6 = |3k|$$|k| = 2$Therefore,$k = \pm 2$.Thus,the correct option is $D$.

Class 12 Mathematics 3-4. Matrices and Determinants Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If the area of a triangle is $3$ sq. units whose vertices are $A(1, 3)$,$B(0, 0)$,and $C(k, 0)$,then $k$ is equal to:

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$2$
B
$-2$
C
$0$
D
$\pm 2$

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3-4. Matrices and Determinants

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Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If the area of a triangle is $3$ sq. units whose vertices are $A(1, 3)$,$B(0, 0)$,and $C(k, 0)$,then $k$ is equal to:

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If $P, Q$ and $R$ are angles of $\Delta PQR$, then the value of $\left|\begin{array}{ccc}-1 & \cos R & \cos Q \\ \cos R & -1 & \cos P \\ \cos Q & \cos P & -1\end{array}\right|$ is equal to

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