The area of a triangle with vertices (K, 0), | vedclass

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

Vedclass · Accepted Answer

$0$ or $8$

Vedclass · Answer

(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by the formula: $	ext{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 $(K, 0), (4, 0), (0, 2)$ and $	ext{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: $-2K + 8 = 8$ or $-2K + 8 = -8$ Case $1$: $-2K = 0 \Rightarrow K = 0$. Case $2$: $-2K = -16 \Rightarrow K = 8$. Thus,the value of $K$ is $0$ or $8$.

The area of a triangle with vertices $(K, 0), (4, 0), (0, 2)$ is $4$ square units. Find the value of $K$.

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