(A) 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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(A) 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 $(3, 5), (2, 2)$ and $(k, 2)$ and $\text{Area} = 3$. Substituting the values: $3 = \frac{1}{2} |3(2 - 2) + 2(2 - 5) + k(5 - 2)|$ $3 = \frac{1}{2} |3(0) + 2(-3) + k(3)|$ $3 = \frac{1}{2} |0 - 6 + 3k|$ $6 = |-6 + 3k|$ This implies two cases: $1) -6 + 3k = 6 \implies 3k = 12 \implies k = 4$ $2) -6 + 3k = -6 \implies 3k = 0 \implies k = 0$ Thus,the values of $k$ are $0$ and $4$.

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

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The area of the triangle whose vertices are $(3,5), (2,2)$ and $(k, 2)$ is $3$ sq. unit. Then,the value of $k$ is . . . . . . .

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A
$0, 4$
B
$0, -4$
C
$3, 1$
D
$-3, 1$

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3 and 4 .Determinants and Matrices

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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 $n \ne 3k$ and $1, \omega, \omega^2$ are the cube roots of unity,then $\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^{2n} & 1 & \omega^n \\ \omega^n & \omega^{2n} & 1 \end{vmatrix}$ has the value

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The area of the triangle whose vertices are $(3,5), (2,2)$ and $(k, 2)$ is $3$ sq. unit. Then,the value of $k$ is . . . . . . .

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