Find the equation of the straight lines passing through the origin making an angle `alpha` with the straight line `y=mx+c`.

Options

A. `(y-y_(1)) = tan (theta - alpha) (x - x_(1))`, where `tan theta = m`
B. `(y-y_(1)) = tan (theta + alpha) (x - x_(1))`, where `tan theta = m`
C. `(y-y_(1)) = tan (theta - alpha) (x + x_(1))`, where `tan theta = m`
D. `(y-y_(1)) = tan (theta + alpha) (x + x_(1))`, where `tan theta = m`

Answer

To find the equation of the straight lines passing through the origin and making an angle \(\alpha\) with the straight line \(y = mx + c\), we can follow these steps:

### Step 1: Identify the slope of the given line
The slope of the line \(y = mx + c\) is \(m\).

### Step 2: Use the angle formula
The tangent of the angle \(\alpha\) between two lines with slopes \(m\) and \(m'\) is given by the formula:
\[ \tan(\alpha) = \left| \frac{m - m'}{1 + mm'} \right| \]
Here, \(m'\) is the slope of the line we want to find.

### Step 3: Set up the equation
From the formula, we can set up two equations because of the absolute value:
1. \(\tan(\alpha) = \frac{m - m'}{1 + mm'}\)
2. \(\tan(\alpha) = \frac{m' - m}{1 + mm'}\)

### Step 4: Solve for \(m'\) from the first equation
Starting with the first equation:
\[ \tan(\alpha)(1 + mm') = m - m' \]
Rearranging gives:
\[ \tan(\alpha) + \tan(\alpha)mm' = m - m' \]
\[ m' + \tan(\alpha)mm' = m - \tan(\alpha) \]
Factoring out \(m'\):
\[ m'(1 + \tan(\alpha)m) = m - \tan(\alpha) \]
Thus,
\[ m' = \frac{m - \tan(\alpha)}{1 + \tan(\alpha)m} \]

### Step 5: Solve for \(m'\) from the second equation
Now, using the second equation:
\[ \tan(\alpha)(1 + mm') = m' - m \]
Rearranging gives:
\[ \tan(\alpha) + \tan(\alpha)mm' = m' - m \]
\[ m' - \tan(\alpha)mm' = m + \tan(\alpha) \]
Factoring out \(m'\):
\[ m'(1 - \tan(\alpha)m) = m + \tan(\alpha) \]
Thus,
\[ m' = \frac{m + \tan(\alpha)}{1 - \tan(\alpha)m} \]

### Step 6: Write the equations of the lines
Now we have two slopes \(m' = \frac{m - \tan(\alpha)}{1 + \tan(\alpha)m}\) and \(m' = \frac{m + \tan(\alpha)}{1 - \tan(\alpha)m}\). The equations of the lines passing through the origin are:
1. \(y = \frac{m - \tan(\alpha)}{1 + \tan(\alpha)m}x\)
2. \(y = \frac{m + \tan(\alpha)}{1 - \tan(\alpha)m}x\)

### Final Answer
Thus, the equations of the straight lines passing through the origin making an angle \(\alpha\) with the line \(y = mx + c\) are:
\[ y = \frac{m - \tan(\alpha)}{1 + \tan(\alpha)m}x \quad \text{and} \quad y = \frac{m + \tan(\alpha)}{1 - \tan(\alpha)m}x \]

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