c) [tex]Ax + By + C = 0[/tex] in normal form is:

[tex]\[
\begin{array}{l}
Ax + By + C = 0 \text{ in normal form } \\
-\frac{A}{\sqrt{A^2 + B^2}} x - \frac{B}{\sqrt{A^2 + B^2}} y = \frac{C}{\sqrt{A^2 + B^2}} \quad (C \ \textgreater \ 0) \\
\text{or, } \frac{A}{\sqrt{A^2 + B^2}} x + \frac{B}{\sqrt{A^2 + B^2}} y = -\frac{C}{\sqrt{A^2 + B^2}} \quad (C \ \textgreater \ 0)
\end{array}
\][/tex]



Answer :

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]