Normal form of line $x+y+1=0$ is

$x \cos \left(45^{\circ}\right)+y \sin \left(135^{\circ}\right)+\frac{1}{\sqrt{2}}=0$

$x \cos \left(45^{\circ}\right)+y \sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

$x \cos \left(45^{\circ}\right)+y \sin \left(45^{\circ}\right)+\frac{1}{\sqrt{2}}=0$

JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

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Application +4 | -1

Normal form of line

x+y+1=0

View answer and explanation

Correct option: C

x \cos \left(225^{\circ}\right)+y \sin \left(225^{\circ}\right)=\frac{1}{\sqrt{2}}

Option C,

x \cos{(225°)} + y \sin{(225°)} = \frac{{1}}{{\sqrt{{2}}}}

is the correct normal form because it represents the given line with the correct orientation and distance from the origin.

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(3,8)

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