Put the equation `12y=5x+65` in the form `x"cos"theta+y"sin"theta=p` and indicate clearly, in a rough diagram the position of the straight line and the meaning of the constant `theta` and p.

Answer

To convert the equation \( 12y = 5x + 65 \) into the form \( x \cos \theta + y \sin \theta = p \), we will follow these steps:

### Step 1: Rearrange the given equation
Start with the equation:
\[ 12y = 5x + 65 \]
Rearranging gives:
\[ 5x - 12y + 65 = 0 \]

### Step 2: Identify coefficients
From the rearranged equation \( 5x - 12y + 65 = 0 \), we can identify:
- \( a = 5 \)
- \( b = -12 \)
- \( c = 65 \)

### Step 3: Convert to the standard form
We need to express the equation in the form \( ax + by + c = 0 \). We already have it in this form:
\[ 5x - 12y + 65 = 0 \]

### Step 4: Find \( p \)
The value of \( p \) is given by the formula:
\[ p = -\frac{c}{\sqrt{a^2 + b^2}} \]
Calculating \( a^2 + b^2 \):
\[ a^2 + b^2 = 5^2 + (-12)^2 = 25 + 144 = 169 \]
Thus, \( \sqrt{a^2 + b^2} = \sqrt{169} = 13 \).

Now substituting \( c \):
\[ p = -\frac{65}{13} = -5 \]

### Step 5: Find \( \cos \theta \) and \( \sin \theta \)
Using the formulas:
\[ \cos \theta = \frac{a}{\sqrt{a^2 + b^2}} = \frac{5}{13} \]
\[ \sin \theta = \frac{b}{\sqrt{a^2 + b^2}} = \frac{-12}{13} \]

### Step 6: Write the final equation
Now we can write the equation in the desired form:
\[ x \cos \theta + y \sin \theta = p \]
Substituting the values we found:
\[ x \cdot \frac{5}{13} + y \cdot \frac{-12}{13} = -5 \]
Multiplying through by 13 to eliminate the fraction:
\[ 5x - 12y = -65 \]

### Step 7: Diagram and interpretation
To visualize the line represented by the equation, we can sketch the coordinate axes and plot the line. The line will intersect the y-axis at \( y = \frac{65}{12} \) and the x-axis at \( x = 13 \).

- **Meaning of \( \theta \)**: The angle \( \theta \) is the angle between the positive x-axis and the line perpendicular to the line represented by the equation.
- **Meaning of \( p \)**: The constant \( p \) represents the perpendicular distance from the origin to the line.