Question

transform the equation 3x+4y=5 in normal form​

Answer 1

Answer:

Let ax+by=c be the equation of a straight line. Here c must be a positive integer. To find its normal form, we have to just divide each term by √(a²+b²).

Given equation 3x+4y=5

Here a=3 and b=4

So √(a²+b²) = √(3²+4²) = √(9+16) = √(25) = 5

Dividing each term by 5 we get

3x/5 + 4y/5 = 1

→ cos[acos(3/5)] + sin[asin(4/5)] = 1

→ cos[atan(4/3)] + sin[atan(4/3)] = 1

Which is normal form of the given equation.

Comparing this equation with cosα + sinα = p

We get,

α = atan(4/3) and p = 1

Hope it will help you.