Question

transform the equation 3x+4y+12=0 into normal form

Answer 1

Answer:

Step-by-step explanation:

To transform any linear equation of the form ax + by + c = 0 into normal form,

follow these steps

1) ax + by = -c , if (c < 0 )

   -ax - by = c , if (c >0)

2)Let me assume  c>0 which we can always make it by multiplying equation by -1,

Now, divide the equation on both sides by √a²+b², so that we get

-ax/√a²+b² - by/√a²+b² = c/√a²+b²,

Now, let p = c/√a²+b² will be the perpendicular distance  from the origin to the line.

If we set , cosФ = -a/√a²+b²   sinФ=-b/√a²+b², then equation turns out to be in the form xcosФ  + ysinФ = p, where Ф turns out to be the angle made by the perpendicular from the origin to the given line and the x-axis.

Now, 3x +4y + 12 = 0,

we write as -3x -4y = 12

Dividing on both sides by √(-3)²+(-4)² = 5, we get

-3x/5 -4y/5 = 12/5

which is in the form

xcosФ  + ysinФ = p,

where cosФ = -3/5, sinФ = -4/5 and p = 12/5.