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प्रश्न
Reduce the equation \[\sqrt{3}\] x + y + 2 = 0 to the normal form and find p and α.
उत्तर
\[\sqrt{3}\] x + y + 2 = 0
\[\Rightarrow - \sqrt{3}x - y = 2\]
\[ \Rightarrow \frac{- \sqrt{3}x}{\sqrt{\left( - \sqrt{3} \right)^2 + \left( - 1 \right)^2}} - \frac{y}{\sqrt{\left( - \sqrt{3} \right)^2 + \left( - 1 \right)^2}} = \frac{2}{\sqrt{\left( - \sqrt{3} \right)^2 + \left( - 1 \right)^2}} \left[ \text { Dividing both sides by } \sqrt{\left( \text { coefficient of } x \right)^2 + \left( \text { coefficient of }y \right)^2} \right]\]
\[ \Rightarrow \frac{- \sqrt{3}x}{2} - \frac{y}{2} = 1\]
This is the normal form of the given line.
Here, p = 1,
\[cos\alpha = - \frac{\sqrt{3}}{2}\] and \[sin\alpha = - \frac{1}{2}\]
\[\Rightarrow \alpha = {210}^\circ\]
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