Reduce the following equation to the normal form and find p and α in \[x + y + \sqrt{2} = 0\].

#### Solution

\[x + y + \sqrt{2} = 0\]

\[\Rightarrow - x - y = \sqrt{2}\]

\[ \Rightarrow - \frac{x}{\sqrt{\left( - 1 \right)^2 + \left( - 1 \right)^2}} - \frac{y}{\sqrt{\left( - 1 \right)^2 + \left( - 1 \right)^2}} = \frac{\sqrt{2}}{\sqrt{\left( - 1 \right)^2 + \left( - 1 \right)^2}} \left[\text { Dividing both sides by } \sqrt{\left( \text { coefficient of x } \right)^2 + \left( \text { coefficient of y } \right)^2} \right]\]

\[ \Rightarrow - \frac{x}{\sqrt{2}} - \frac{y}{\sqrt{2}} = 1\]

This is the normal form of the given line, where p = 1,

\[cos\alpha = - \frac{1}{\sqrt{2}}\]

\[\sin\alpha = - \frac{1}{\sqrt{2}}\]

\[ \Rightarrow \alpha = {225}^\circ \left[ \because \text { The coefficent of x and y are negative . So, } \alpha \text { lies in third quadrant } \right]\]