Question

Normal form of the line x+y+√2=0is

Answer 1

Step-by-step explanation:

Hope this will help .....

Answer 2

Answer:

Step-by-step explanation:

Concept:

The equation for the line is provided by $x cos \alpha +y sin \alpha = p$ where the length of the perpendicular from the origin is p and the angle created by the perpendicular with the positive x-axis is given by. This is referred to as the line's typical form.

Given:

The equation $x+y+\sqrt{2} =0$

Find:

To find the normal form of the line $x+y+\sqrt{2} =0$

Solution:

We have to find the normal form of the line

$x + y +\sqrt{2} = 0 ...(i)$

We know the normal form of a line (i) is

$x cos\alpha + y cos\alpha = p .....(ii)$ ; where $p > 0$

[$\alpha$ be the slope of  (I) and $p$be the perpendicular distance to the (I) from origin

Since the equation (I) and (ii) represents the same , so we get

⇒ $cos \alpha /1 = sin\alpha /1 = p/\sqrt{2}$

⇒ $cos\alpha / -1 = sin\alpha /-1 = p/+\sqrt{2}$ $= \sqrt{cos^{2}\alpha + sin^{2}\alpha} / \sqrt{1^{2} + 1^{2} } = 1/\sqrt{2}$

⇒$cos\alpha = -1/\sqrt{2} , sin\alpha = -1/\sqrt{2} , p=1$

Since $cos\alpha$ and $sin\alpha$ are both negative, so $\alpha$ lis in $3rd$coordinates

∴$\alpha = 3\pi /4$ as $cos = 3\pi /4 = sin\3pi / 4 = -1/\sqrt{2}$

Hence, Normal form of line (I) is

$x cos (3\pi /4) + y sin (3\pi /4) = 1$

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