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To find the general equation of a line in \( \mathbb{R}^2 \) passing through the point \( (1,3) \) and parallel to the vector \( \begin{pmatrix} 2 \\ -7 \end{pmatrix} \), we can follow these steps:

### Step 1: Determine the Direction Vector
The line is parallel to \( \begin{pmatrix} 2 \\ -7 \end{pmatrix} \), which means the direction vector \( \vec{d} \) for our line is \( \vec{d} = \begin{pmatrix} 2 \\ -7 \end{pmatrix} \).

### Step 2: Formulate the Slope-Intercept Form
To use the vector \( \begin{pmatrix} 2 \\ -7 \end{pmatrix} \), we can convert it into a slope. The slope \( m \) of the line is given by:
\[
m = \frac{\text{rise}}{\text{run}} = \frac{-7}{2}
\]

Thus, the slope of our line is \( -\frac{7}{2} \).

### Step 3: Use Point-Slope Form of the Line Equation
Since we know the point \( (1,3) \) and the slope \( -\frac{7}{2} \), we can use the point-slope form:
\[
y - y_1 = m(x - x_1)
\]
Substitute \( (x_1, y_1) = (1, 3) \) and \( m = -\frac{7}{2} \):
\[
y - 3 = -\frac{7}{2}(x - 1)
\]

### Step 4: Simplify to the Standard Form
Expanding this equation:
\[
y - 3 = -\frac{7}{2}x + \frac{7}{2}
\]
Multiply everything by 2 to clear the fraction:
\[
2y - 6 = -7x + 7
\]
Rearrange to standard form \( Ax + By = C \):
\[
7x + 2y = 13
\]

### Conclusion
The equation of the line is:
\[
7x + 2y = 13
\]

Thus, the correct answer is:
**A: \( 7x + 2y = 13 \)**

Would you like more details or have any questions?

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Here are some related questions to further explore this topic:
1. How would the equation change if the line were perpendicular to \( \begin{pmatrix} 2 \\ -7 \end{pmatrix} \) instead?
2. What is the significance of the direction vector when determining the line's equation?
3. How do you convert from point-slope form to standard form in general?
4. What would be the steps if the line had to pass through a different point, say \( (2, 5) \)?
5. How does one find the equation of a line in 3D space given a point and a direction vector?

**Tip:** When a line is parallel to a given vector, the direction vector's components can directly provide the slope when written as a fraction of rise over run.

Find the general equation of a line in \( \mathbb{R}^2 \) passing through the point \( (1,3) \) and parallel to \( \begin{pmatrix} 2 \\ -7 \end{pmatrix} \).

Learn how to find the general equation of a line passing through the point (1,3) in R^2 that is parallel to the vector (2, -7). This solution involves linear equations, point-slope form, and parallel line concepts. Ideal for high school math students.

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