Question

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

Answer 1

Answer:

x + y +4 = 0

Step-by-step explanation:

Given,

$p = 2\sqrt{2}$  $and$  $\alpha = \frac{5\pi }{4}$

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

So,we have,

$x cos(\alpha ) + ysin(\alpha ) = p\\\\=> x cos(\frac{5\pi }{4}) + ysin\frac{5\pi }{4} = 2\sqrt{2} \\\\=> x cos(\pi +\frac{\pi }{4}) + ysin(\pi +\frac{\pi }{4}) = 2\sqrt{2}\\\\=>-x cos(\frac{\pi }{4} ) - ysin(\frac{\pi }{4}) = 2\sqrt{2}\\\\=> - \frac{x}{\sqrt{2}} - \frac{y}{\sqrt{2}} = 2\sqrt{2}$

Multiplying both side by √2, we get,

$=> -x-y = 4 \\\\=>x + y +4 = 0$

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