the image of the point `(-1,3,4)` in the plane `x-2y=0` a.`(-(17)/(3),(19)/(3),4)` b.(15,11,4) c.`(-(17)/(3),(19)/(3),1)` d.`((9)/(5),-(13)/(5),4)`

Options

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

Explanation

To find the image of the point \((-1, 3, 4)\) with respect to the plane defined by the equation \(x - 2y = 0\), we can use the formula for the image of a point with respect to a plane.

### Step 1: Identify the coefficients of the plane
The equation of the plane can be rewritten in the standard form \(Ax + By + Cz = D\). Here, we can express it as:
\[ 1 \cdot x + (-2) \cdot y + 0 \cdot z = 0 \]
Thus, we have:
- \(A = 1\)
- \(B = -2\)
- \(C = 0\)
- \(D = 0\)

### Step 2: Identify the coordinates of the point
The coordinates of the point are:
\[ (x_1, y_1, z_1) = (-1, 3, 4) \]

### Step 3: Calculate the value of \(Ax_1 + By_1 + Cz_1\)
Now, we calculate:
\[ Ax_1 + By_1 + Cz_1 = 1 \cdot (-1) + (-2) \cdot 3 + 0 \cdot 4 = -1 - 6 + 0 = -7 \]

### Step 4: Calculate \(A^2 + B^2 + C^2\)
Next, we calculate:
\[ A^2 + B^2 + C^2 = 1^2 + (-2)^2 + 0^2 = 1 + 4 + 0 = 5 \]

### Step 5: Use the formula to find the image
The formula for the image of the point \((x_1, y_1, z_1)\) with respect to the plane is given by:
\[ \left( x - x_1, y - y_1, z - z_1 \right) = -2 \cdot \frac{Ax_1 + By_1 + Cz_1}{A^2 + B^2 + C^2} \cdot (A, B, C) \]
Substituting the values we have:
\[ \left( x + 1, y - 3, z - 4 \right) = -2 \cdot \frac{-7}{5} \cdot (1, -2, 0) \]
Calculating the right-hand side:
\[ \left( x + 1, y - 3, z - 4 \right) = \frac{14}{5} \cdot (1, -2, 0) = \left( \frac{14}{5}, -\frac{28}{5}, 0 \right) \]

### Step 6: Solve for \(x\), \(y\), and \(z\)
Now, we can write the equations:
1. \(x + 1 = \frac{14}{5}\)
2. \(y - 3 = -\frac{28}{5}\)
3. \(z - 4 = 0\)

From the first equation:
\[ x = \frac{14}{5} - 1 = \frac{14}{5} - \frac{5}{5} = \frac{9}{5} \]

From the second equation:
\[ y = -\frac{28}{5} + 3 = -\frac{28}{5} + \frac{15}{5} = -\frac{13}{5} \]

From the third equation:
\[ z = 4 \]

### Step 7: Final coordinates of the image
Thus, the image of the point \((-1, 3, 4)\) in the plane \(x - 2y = 0\) is:
\[ \left( \frac{9}{5}, -\frac{13}{5}, 4 \right) \]

### Answer
The correct option is:
**d. \(\left( \frac{9}{5}, -\frac{13}{5}, 4 \right)\)**