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Question
Reduce the following equation to the normal form and find p and α in \[x - y + 2\sqrt{2} = 0\].
Solution
\[x - y + 2\sqrt{2} = 0\]
\[\Rightarrow - x + y = 2\sqrt{2}\]
\[ \Rightarrow - \frac{x}{\sqrt{\left( - 1 \right)^2 + \left( 1 \right)^2}} + \frac{y}{\sqrt{\left( - 1 \right)^2 + \left( 1 \right)^2}} = \frac{2\sqrt{2}}{\sqrt{\left( - 1 \right)^2 + \left( 1 \right)^2}} \left[ \text { Dividing both sides by } \sqrt{\left(\text { coefficient of x } \right)^2 + \left(\text { coefficient of y } \right)^2} \right]\]
\[ \Rightarrow - \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 2\]
This is the normal form of the given line, where p = 2,
\[cos\alpha = - \frac{1}{\sqrt{2}}\] and \[\sin\alpha = \frac{1}{\sqrt{2}}\]
\[ \Rightarrow \alpha = {135}^\circ \left[ \because \text { The coefficent of x and y are negative and positive respectively . So }, \alpha \text { lies in second quadrant } \right]\]
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