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# Find values of $k$ if area of triangle is $4\; sq. units$ and vertices are $(-2, 0), (0, 4), (0, k)$

$\begin{array}{1 1} 1 \;and\; -8 \\ 1\; and\; 8 \\ 0\; and\; -8 \\ 0\; and\; 8 \end{array}$

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A)
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• The area of a triangle whose vertices are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by
• $\bigtriangleup=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}$
• $|\bigtriangleup|=\frac{1}{2}\begin{bmatrix}x_1\begin{vmatrix}y_2 & 1\\y_3 & 1\end{vmatrix}-y_1\begin{vmatrix}x_2 & 1\\x_3 & 1\end{vmatrix}+1\begin{vmatrix}x_2 & y_2\\x_3 & y_3\end{vmatrix}\end{bmatrix}$
Let $(-2,0)=(x_1,y_1);(0,4)=(x_2,y_2);(0,k)=(x_3,y_3)$
The area of the triangle is given by
$\bigtriangleup=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}$
Now substituting the respective values we get
Given area =4sq.units.
Hence $4=\frac{1}{2}\begin{vmatrix}-2 & 0 & 1\\0 & 4 & 1\\0 & k & 1\end{vmatrix}$
Now expanding along the row $R_1$ we get
$\mid\bigtriangleup\mid=\frac{1}{2}\begin{bmatrix}-2\begin{vmatrix}4 & 1\\k & 1\end{vmatrix}-0\begin{vmatrix}0 & 1\\0 & 1\end{vmatrix}+1\begin{vmatrix}0 & 4\\0 & k\end{vmatrix}\end{bmatrix}$
We take + or - because the determinant value is absolute value and hence it can be either + or -.
8=$\pm[-2(4-k)]$
Therefore if 8=+(-8+2k)
then 2k=16
k=8
Therefore if 8=-(-8+2k)
then k=8-8
k=0.
Hence the value of k is (0,8)