Determine the equation of the line that passes through the point T(1, -2), and with the positive part of the x axis closes an angle twice as large as the line 9x - 3y + 11 = 0

To determine the equation of the line that passes through the point \( T(1, -2) \) and closes an angle with the positive part of the x-axis that is twice as large as the angle closed by the line \( 9x - 3y + 11 = 0 \), we can follow these steps:

### Step 1: Find the slope of the given line.
The equation of the line is given as \( 9x - 3y + 11 = 0 \). To find the slope, we need to express this equation in slope-intercept form, \( y = mx + c \).

\[
9x - 3y + 11 = 0
\]

\[
-3y = -9x - 11
\]

\[
y = 3x + \frac{11}{3}
\]

Here, the slope \( m_1 \) of the given line is 3.

### Step 2: Determine the angle with the x-axis.
The slope \( m_1 \) of a line is related to the angle \( \theta_1 \) that the line makes with the positive x-axis by the following relation:

\[
\tan(\theta_1) = m_1 = 3
\]

So,

\[
\theta_1 = \tan^{-1}(3)
\]

### Step 3: Find the angle for the new line.
Let the angle that the required line makes with the positive x-axis be \( \theta_2 \). According to the problem, \( \theta_2 = 2\theta_1 \).

### Step 4: Find the slope of the new line.
We know that:

\[
\tan(\theta_2) = \tan(2\theta_1)
\]

Using the double-angle formula for tangent:

\[
\tan(2\theta_1) = \frac{2\tan(\theta_1)}{1 - \tan^2(\theta_1)}
\]

Substituting \( \tan(\theta_1) = 3 \):

\[
\tan(2\theta_1) = \frac{2 \times 3}{1 - 3^2} = \frac{6}{1 - 9} = \frac{6}{-8} = -\frac{3}{4}
\]

So, the slope \( m_2 \) of the required line is \( -\frac{3}{4} \).

### Step 5: Use the point-slope form to find the equation of the line.
The line passes through the point \( T(1, -2) \) and has a slope \( m_2 = -\frac{3}{4} \). The point-slope form of the equation of a line is given by:

\[
y - y_1 = m(x - x_1)
\]

Substituting \( (x_1, y_1) = (1, -2) \) and \( m = -\frac{3}{4} \):

\[
y + 2 = -\frac{3}{4}(x - 1)
\]

Multiply both sides by 4 to eliminate the fraction:

\[
4(y + 2) = -3(x - 1)
\]

\[
4y + 8 = -3x + 3
\]

\[
3x + 4y = -5
\]

So, the equation of the required line is:

\[
\boxed{3x + 4y + 5 = 0}
\]

Would you like any details or have any questions?  

Here are 5 related questions for further exploration:

1. How can the angle of inclination of a line be derived from its slope?
2. What is the geometric interpretation of the tangent function in the context of a line’s slope?
3. How do different transformations affect the slope of a line?
4. How can the equation of a line be determined if two points on the line are known?
5. How would the problem change if the angle was three times as large instead of twice?

**Tip:** When working with slopes, remember that positive slopes indicate lines that rise as they move from left to right, while negative slopes indicate lines that fall.

Learn how to find the equation of a line passing through point T(1, -2) with an angle relation with another line. This problem involves geometry and trigonometry concepts, including slopes, angles with the x-axis, and the double-angle formula for tangent. Suitable for Grades 10-12 students.

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