The image of the point `(-1, 3, 4)` in the plane `x-2y=0` is

Options

A. `(15, 11, 4)`
B. `(-(17)/(3), -(19)/(3), 1)`
C. `(8, 4, 4)`
D. `((9)/(5), (-13)/(5), 4)`

Answer

To find the image of the point \((-1, 3, 4)\) in the plane defined by the equation \(x - 2y = 0\), we will follow these steps:

### Step 1: Identify the given point and the plane
The given point is \(A(-1, 3, 4)\) and the equation of the plane is \(x - 2y = 0\).

### Step 2: Find the foot of the perpendicular from point A to the plane
To find the foot of the perpendicular \(P\) from point \(A\) to the plane, we need to derive the parametric equations of the line passing through \(A\) and perpendicular to the plane. The normal vector to the plane \(x - 2y = 0\) is given by the coefficients of \(x\) and \(y\), which is \((1, -2, 0)\).

The parametric equations of the line can be written as:
\[ \begin{align*} x &= -1 + t \\ y &= 3 - 2t \\ z &= 4 \end{align*} \]

### Step 3: Substitute the line equations into the plane equation
Substituting the parametric equations into the plane equation \(x - 2y = 0\):
\[ (-1 + t) - 2(3 - 2t) = 0 \]
Simplifying this:
\[ -1 + t - 6 + 4t = 0 \\ 5t - 7 = 0 \\ t = \frac{7}{5} \]

### Step 4: Find the coordinates of point P
Now, substituting \(t = \frac{7}{5}\) back into the parametric equations to find the coordinates of point \(P\):
\[ \begin{align*} x_P &= -1 + \frac{7}{5} = \frac{2}{5} \\ y_P &= 3 - 2 \cdot \frac{7}{5} = 3 - \frac{14}{5} = \frac{1}{5} \\ z_P &= 4 \end{align*} \]
Thus, the coordinates of point \(P\) are \(\left(\frac{2}{5}, \frac{1}{5}, 4\right)\).

### Step 5: Find the image point B
Since \(P\) is the midpoint of segment \(AB\), we can use the midpoint formula to find the coordinates of point \(B\):
\[ \begin{align*} \frac{x_A + x_B}{2} &= x_P \\ \frac{-1 + x_B}{2} &= \frac{2}{5} \\ x_B &= 2 \cdot \frac{2}{5} + 1 = \frac{4}{5} + 1 = \frac{9}{5} \end{align*} \]

For the \(y\) coordinate:
\[ \begin{align*} \frac{y_A + y_B}{2} &= y_P \\ \frac{3 + y_B}{2} &= \frac{1}{5} \\ y_B &= 2 \cdot \frac{1}{5} - 3 = \frac{2}{5} - 3 = \frac{2}{5} - \frac{15}{5} = -\frac{13}{5} \end{align*} \]

For the \(z\) coordinate:
\[ \begin{align*} \frac{z_A + z_B}{2} &= z_P \\ \frac{4 + z_B}{2} &= 4 \\ z_B &= 2 \cdot 4 - 4 = 8 - 4 = 4 \end{align*} \]

### Final Coordinates of Point B
Thus, the coordinates of point \(B\) are:
\[ B\left(\frac{9}{5}, -\frac{13}{5}, 4\right) \]

### Conclusion
The image of the point \((-1, 3, 4)\) in the plane \(x - 2y = 0\) is \(\left(\frac{9}{5}, -\frac{13}{5}, 4\right)\).

---