(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: Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(

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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: 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 $(k, 0), (4, 0)$ and $(0, 2)$ and Area $= 4$. Substituting the values: $4 = \frac{1}{2} |k(0 - 2) + 4(2 - 0) + 0(0 - 0)|$. $4 = \frac{1}{2} |-2k + 8|$. $8 = |-2k + 8|$. This implies two cases: Case $1$: $-2k + 8 = 8 \implies -2k = 0 \implies k = 0$. Case $2$: $-2k + 8 = -8 \implies -2k = -16 \implies k = 8$. Thus,the values of $k$ are $0$ and $8$.

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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If the area of a triangle is $4$ sq. units whose vertices are $(k, 0), (4, 0)$ and $(0, 2)$,then the value of $k$ is . . . . . . .

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$0, 8$
B
$0, -8$
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$0, 16$
D
$0, -16$

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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 the area of a triangle is $4$ sq. units whose vertices are $(k, 0), (4, 0)$ and $(0, 2)$,then the value of $k$ is . . . . . . .

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