Given the equation [tex]\( A x + B y + C = 0 \)[/tex], we need to convert it to its normal form. Let's go through the process step by step.
### Step 1: Identify the coefficients
Here, we have:
- [tex]\( A \)[/tex]
- [tex]\( B \)[/tex]
- [tex]\( C \)[/tex]
### Step 2: Calculate the denominator
The denominator for the conversion to normal form is calculated using the formula [tex]\( \sqrt{A^2 + B^2} \)[/tex].
### Step 3: Convert to normal form
We now convert the given line equation [tex]\( A x + B y + C = 0 \)[/tex] into its normal forms:
#### Normal Form 1
The first normal form is:
[tex]\[ -\frac{A}{\sqrt{A^2 + B^2}} x - \frac{B}{\sqrt{A^2 + B^2}} y = \frac{C}{\sqrt{A^2 + B^2}} \][/tex]
#### Normal Form 2
The second normal form is:
[tex]\[ \frac{A}{\sqrt{A^2 + B^2}} x + \frac{B}{\sqrt{A^2 + B^2}} y = -\frac{C}{\sqrt{A^2 + B^2}} \][/tex]
### Step 4: Applying the values from the calculation
From the provided solution, we have numerical results for this conversion:
- For Normal Form 1, we get:
[tex]\[
\left( -\frac{A}{\sqrt{A^2 + B^2}}, -\frac{B}{\sqrt{A^2 + B^2}}, \frac{C}{\sqrt{A^2 + B^2}} \right) = \left( -0.7071067811865475, -0.7071067811865475, 0.7071067811865475 \right)
\][/tex]
- For Normal Form 2, we get:
[tex]\[
\left( \frac{A}{\sqrt{A^2 + B^2}}, \frac{B}{\sqrt{A^2 + B^2}}, -\frac{C}{\sqrt{A^2 + B^2}} \right) = \left( 0.7071067811865475, 0.7071067811865475, -0.7071067811865475 \right)
\][/tex]
Therefore, for the given [tex]\( A = 1 \)[/tex], [tex]\( B = 1 \)[/tex], and [tex]\( C = 1 \)[/tex], the equation [tex]\( 1x + 1y + 1 = 0 \)[/tex] in normal form is:
[tex]\[
-\frac{1}{\sqrt{1^2 + 1^2}} x - \frac{1}{\sqrt{1^2 + 1^2}} y = \frac{1}{\sqrt{1^2 + 1^2}}
\][/tex]
or
[tex]\[
\frac{1}{\sqrt{1^2 + 1^2}} x + \frac{1}{\sqrt{1^2 + 1^2}} y = -\frac{1}{\sqrt{1^2 + 1^2}}
\][/tex]
Which simplifies numerically to:
[tex]\[
-0.7071067811865475 x - 0.7071067811865475 y = 0.7071067811865475
\][/tex]
or
[tex]\[
0.7071067811865475 x + 0.7071067811865475 y = -0.7071067811865475
\][/tex]
