Question

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

Answer 1

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

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

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

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

This is the normal form of the given line, where p = 2,

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

\[ \Rightarrow \alpha = {135}^\circ \left[ \because \text { The coefficent of x and y are negative and positive respectively . So }, \alpha \text { lies in second quadrant } \right]\]