Hello, Kataangel!
The area of the triangle bounded by the three lines: .$\displaystyle \begin{Bmatrix}x \:=\:0 \\ y\:=\:0 \\ x+y\:=\:a\end{Bmatrix}$ has an area of 8.
What is $\displaystyle a$ ? Did you make a sketch? Code:

* 
a*
:*
:::*
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  +      *  
0 a *
$\displaystyle \text{The line }x = 0\text{ is the }y\text{axis.}$
$\displaystyle \text{The line }y = 0 \text{ is the }x\text{axis.}$
$\displaystyle \text{The line }x+y\:=\:0\text{ has intercepts: }\,(a,0),\:(0,a)$
We have a right triangle with: .$\displaystyle \text{base} = a,\;\text{height} = a$
The area of a triangle is: .$\displaystyle \text{Area} \:=\:\frac{1}{2}\text{(base)(height)}$
So we have: .$\displaystyle \frac{1}{2}(a)(a) \:=\:8 \quad\Rightarrow\quad a^2 \:=\:16 \quad\Rightarrow\quad \boxed{a \:=\:\pm4}$