Question

find the equation of straight line with the normal form p=2√2,a= t 5π/4​

Answer 1

Given,

p = 2\sqrt{2}p=22 andand \alpha = \frac{5\pi }{4}α=45π

Here, 225 degree=(180+45) degree=(π+(π/4))=(5π)/4

So,we have,

\begin{gathered}x cos(\alpha ) + ysin(\alpha ) = p\\\\= > x cos(\frac{5\pi }{4}) + ysin\frac{5\pi }{4} = 2\sqrt{2} \\\\= > x cos(\pi +\frac{\pi }{4}) + ysin(\pi +\frac{\pi }{4}) = 2\sqrt{2}\\\\= > -x cos(\frac{\pi }{4} ) - ysin(\frac{\pi }{4}) = 2\sqrt{2}\\\\= > - \frac{x}{\sqrt{2}} - \frac{y}{\sqrt{2}} = 2\sqrt{2}\end{gathered}xcos(α)+ysin(α)=p=>xcos(45π)+ysin45π=22=>xcos(π+4π)+ysin(π+4π)=22=>−xcos(4π)−ysin(4π)=22=>−2x−2y=22

Multiplying both side by √2, we get,

\begin{gathered}= > -x-y = 4 \\\\= > x + y +4 = 0\end{gathered}=>−x−y=4=>x+y+4=0

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