The equation of a plane through the point (2, 3, 1)
and (4, -5, 3) and parallel to x- axis

Options

A. `x + y + 4z = 7`
B. `y+ 4z = 7`
C. `y- 4z = 7`
D. `y+ 4x = 7`

Answer

To find the equation of a plane that passes through the points (2, 3, 1) and (4, -5, 3) and is parallel to the x-axis, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Plane's Orientation**:
Since the plane is parallel to the x-axis, it means that for any point on the plane, the x-coordinate can take any value, while the y and z coordinates remain constant.

2. **Identifying the Points**:
We have two points through which the plane passes:
- Point A: (2, 3, 1)
- Point B: (4, -5, 3)

3. **Finding the Direction Vector**:
The direction vector can be calculated using the two points:
\[ \text{Direction Vector} = B - A = (4 - 2, -5 - 3, 3 - 1) = (2, -8, 2) \]

4. **Finding the Normal Vector**:
Since the plane is parallel to the x-axis, the normal vector to the plane will be in the direction of the y-z plane. Thus, we can take the normal vector as:
\[ \text{Normal Vector} = (0, 1, 4) \]
This is because the change in x is 0 (as the plane is parallel to the x-axis), and we can derive the coefficients for y and z from the points.

5. **Using the Point-Normal Form of the Plane**:
The equation of the plane can be expressed using the point-normal form:
\[ 0(x - x_0) + 1(y - y_0) + 4(z - z_0) = 0 \]
Using point A (2, 3, 1):
\[ 0(x - 2) + 1(y - 3) + 4(z - 1) = 0 \]
Simplifying this gives:
\[ y - 3 + 4z - 4 = 0 \]
Rearranging this leads to:
\[ y + 4z = 7 \]

6. **Final Equation of the Plane**:
The equation of the plane is:
\[ y + 4z = 7 \]

### Conclusion:
The equation of the plane that passes through the points (2, 3, 1) and (4, -5, 3) and is parallel to the x-axis is:
\[ y + 4z = 7 \]