The area of triangle with vertices \( (\mathrm{K}, 0),(4,0),(0,2) \) is \( 4 \) square units, then value of \( \mathrm{K} \) is

Given vertices of a triangle is, \( (\mathrm{K}, 0),(4,0),(0,2) \) and area of triangle is \( 4 \mathrm{sq} \). units.
We know that, area of triangle with vertices \( \left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\left(x_{3}, y_{3}\right) \) is given by
\[
\begin{aligned}
A=&\left|\begin{array}{ccc}
x_{1} & x_{2} & x_{3} \\
y_{1} & y_{2} & y_{3} \\
z_{1} & z_{2} & z_{3}
\end{array}\right| \\
A=& \frac{1}{2} \mid\left(x_{1}\left(y_{1}-y_{3}\right)+x_{2}\left(y_{3}-y_{2}\right)+x_{3}\left(y_{1}-y_{2}\right)\right)
\end{aligned}
\]
So,
\[
\begin{array}{l}
A=\frac{1}{2}|K(0-2)+4(2-0)-0(0-0)|=4 \\
\Rightarrow \frac{1}{2}|-2 K+8|=4 \\
\Rightarrow 8-2 K=\pm 8 \\
=8+8,60=8
\end{array}
\]
For \( +8 \), we have \( 8-2 K=8 \Rightarrow K=0 \)
For \( -8 \), we have \( 8-2 K=-8 \Rightarrow K=8 \)