Question

Transform the equation x+y-2=0 into normal form of the straight line?

Answer 1

Answer:

$\cos(\frac{\pi}{4}) x+\sin(\frac{\pi}{4})  y=\sqrt{2}$

Step-by-step explanation:

We know that if Ax + By = C where  C > 0

then the equation of the line in normal form is given by

$\frac{A}{\sqrt{A^2+B^2}} x+\frac{B}{\sqrt{A^2+B^2}}y=\frac{C}{\sqrt{A^2+B^2}}$

if $\frac{A}{\sqrt{A^2+B^2}} = \cos\alpha$

$\frac{B}{\sqrt{A^2+B^2}} = \sin\alpha$, and

$\frac{C}{\sqrt{A^2+B^2}} = p$

Then the equation becomes

$\cos\alpha x+\sin\alpha x y=p$

Given equation

x + y - 2 = 0

or, x + y = 2

Here, A = 1, B = 1 and C = 2

Dividing the equation by $\sqrt{1^2+1^2}=\sqrt{2}$

$\frac{1}{\sqrt{2} } x+\frac{1}{\sqrt{2} } y=\frac{2}{\sqrt{2} }$

or, $\cos(\frac{\pi}{4}) x+\sin(\frac{\pi}{4})y=\sqrt{2}$

which is the required equation in normal form

Answer 2

Answer:

why should we divide by root2 and take the values A as 1 B as 1