Question

Reduce the equation x+y+√2=0 to the normal form​

Answer 1

Answer:

x+y=( -√2)

Step-by-step explanation:

Answer 2

Answer :

x•cos(3π/4) + y•sin(3π/4) = 1

Working rule :

To convert general form of straight line in its normal form :

Let ax + by + c = 0 be a straight line .

=> -ax - by = c

=> -ax/√(a² + b²) - by/√(a² + b²) = c/√(a² + b²)

=> x·cosα + y·sinα = p

Where cosα = -a/√(a² + b²)

sinα = -b/√(a² + b²)

p = c/√(a² + b²)

Solution :

Here ,

The given equation of straight line is :

x + y + √2 .

Now ,

Comparing the given straight line with general equation ax + by + c = 0 , we have ;

a = 1

b = 1

c = √2

Also ,

=> √(a² + b²) = √(1² + 1²)

=> √(a² + b²) = √(1 + 1)

=> √(a² + b²) = √2

Now ,

=> x + y + √2 = 0

=> -x - y = √2

Now ,

Dividing the above equation by √2 , we get ;

=> -x/√2 - y/√2 = √2/√2

=> x•(-1/√2) + y•(-1/√2) = 1

=> x•cos(π - π/4) + y•sin(π - π/4) = 1

=> x•cos(3π/4) + y•sin(3π/4) = 1

Hence ,

The the normal form of the given equation is :

x•cos(3π/4) + y•sin(3π/4) = 1