با توجه به نقطهٔ داده شده از خط d معادله خط را می‌یابیم:

$a = \frac{{ - 2 - 0}}{{0 - ( - 4)}} = \frac{{ - 2}}{4} =  - \frac{1}{2}$

$y = ax + b = \xrightarrow[{b =  - 2}]{{a =  - \frac{1}{2}}}y =  - \frac{1}{2}x - 2$

با حل دستگاه نقطهٔ برخورد دو خط را می‌یابیم:

$\begin{gathered}
  \left\{ {\begin{array}{*{20}{c}}
  {y =  - \frac{1}{2}x - 2} \\ 
  {2x + y = 1} 
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}
  {y + \frac{1}{2}x =  - 2} \\ 
  {y + 2x = 1} 
\end{array}} \right. \to \underline {\left\{ {\begin{array}{*{20}{c}}
  {y + \frac{1}{2}x =  - 2} \\ 
  { - y - 2x =  - 1} 
\end{array}} \right.}  \hfill \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \frac{3}{2}x =  - 3 \to x = 2 \hfill \\ 
\end{gathered} $

$2(2) + y = 1 \to y =  - 3$

$A\left| {\begin{array}{*{20}{c}}
  2 \\ 
  { - 3} 
\end{array}} \right.$

${x_A} + {y_A} = 2 + ( - 3) =  - 1$