Equation of line in Straight lines

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$The normal form of the line x + y + 1 = 0 is x \cos α + y \sin α = p then α =$

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Solution

$Reduce the equation x + y + 1 = 0 in normal form.$
$Separate the constant term and then divide both sides with \sqrt{a^{2} + b^{2}} = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$
$\begin{array}{l} \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = - \frac{1}{\sqrt{2}} \\ - \frac{x}{\sqrt{2}} - \frac{y}{\sqrt{2}} = \frac{1}{\sqrt{2}} \end{array}$
This can be rewrite as
$x \cos (\frac{5 π}{4}) + y \sin (\frac{5 π}{4}) = \frac{1}{\sqrt{2}}$
$Compare this equation with x \cos α + y \sin α = p$
$Hence, α = \frac{5 π}{4} = 3.93$

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Equation of line in Straight lines

Remember concepts with our Masterclasses.

The normal form of the line x+y+1=0 is xcos⁡α+ysin⁡α=p then α=

Reduce the equation x+y+1=0 in normal form. Separate the constant term and then divide both sides with a2+b2=12+12=2x2+y2=−12−x2−y2=12This can be rewrite as xcos5π4+ysin5π4=12 Compare this equation with xcos⁡α+ysin⁡α=p Hence, α=5π4=3.93

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$The normal form of the line x + y + 1 = 0 is x \cos α + y \sin α = p then α =$