Question

find the equation of straight line with the normal form p=2root2,alfa=5pie/4

Answer 1

Answer:

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Step-by-step explanation:

$given = \\ p = 2 \sqrt{2} \\ \alpha = \frac{5\pi}{4} \\ \\ so \: as \: \: \frac{5\pi}{4} \: \: lies \: in \: quadrant \: 3 \\ \\ values \: of \: sin \: \alpha \: and \: cos \: \alpha \: would \\ be \: negative \: \\ \\ hence \: then \\ \\ sin \: \frac{5\pi}{4} = sin \: (\pi + \frac{\pi}{4} \: ) \\ \\ = sin \: \frac{\pi}{4} \\ \\ = - \: \frac{1}{ \sqrt{2} } \\ \\ similarly \\ \\ cos \: \frac{5\pi}{4} = cos \: \frac{\pi}{4} \\ \\ = - \: \frac{1}{ \sqrt{2} }$

$so \: we \: know \: that \\ normal \: form \: equation \: of \: straight \: \\ line \: is \: given \: by \\ \\ x.cos \: \alpha + y.sin \: \alpha = p \\ \\ \frac{ - 1}{ \sqrt{2} } .x + ( \frac{ - 1}{ \sqrt{2} } .y) = 2 \sqrt{2} \\ \\ \frac{ - x}{ \sqrt{2} } - \frac{y}{ \sqrt{2} } = 2 \sqrt{2} \\ \\ \frac{ - x - y}{ \sqrt{2} } = 2 \sqrt{2} \\ \\ - x - y = 2 \sqrt{2} \times \sqrt{2} \\ \\ - x - y = (2 \times 2) \\ \\ - x - y = 4 \\ \\ x + y + 4 = 0 \\ \\ or \\ \\ \\ x + y = - 4$