Find the distance of the point `(2,1,-1)` is equidistant from the plane `x-2y+4z=9`.

Answer

To find the distance of the point \( (2, 1, -1) \) from the plane given by the equation \( x - 2y + 4z = 9 \), we can use the formula for the distance \( D \) from a point \( (x_0, y_0, z_0) \) to the plane defined by the equation \( Ax + By + Cz + D = 0 \):

\[ D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

### Step 1: Rewrite the plane equation
First, we need to rewrite the plane equation in the form \( Ax + By + Cz + D = 0 \).

The given plane equation is:
\[ x - 2y + 4z = 9 \]
We can rearrange this to:
\[ x - 2y + 4z - 9 = 0 \]
Here, we identify:
- \( A = 1 \)
- \( B = -2 \)
- \( C = 4 \)
- \( D = -9 \)

### Step 2: Identify the point coordinates
The coordinates of the point are:
\[ (x_0, y_0, z_0) = (2, 1, -1) \]

### Step 3: Substitute values into the distance formula
Now we substitute \( A, B, C, D, x_0, y_0, z_0 \) into the distance formula:

\[ D = \frac{|1(2) + (-2)(1) + 4(-1) - 9|}{\sqrt{1^2 + (-2)^2 + 4^2}} \]

### Step 4: Calculate the numerator
Calculating the numerator:
\[ 1(2) + (-2)(1) + 4(-1) - 9 = 2 - 2 - 4 - 9 = -13 \]
Taking the absolute value:
\[ |-13| = 13 \]

### Step 5: Calculate the denominator
Calculating the denominator:
\[ \sqrt{1^2 + (-2)^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21} \]

### Step 6: Final distance calculation
Now we can find the distance \( D \):
\[ D = \frac{13}{\sqrt{21}} \]

Thus, the distance of the point \( (2, 1, -1) \) from the plane \( x - 2y + 4z = 9 \) is:
\[ D = \frac{13}{\sqrt{21}} \]