The length of perpendicular from `P(2-3i)` on the line `(3+4i)Z + (3-4i)bar Z+9=0` is equal to

Answer

To find the length of the perpendicular from the point \( P(2 - 3i) \) to the line given by the equation \( (3 + 4i)Z + (3 - 4i)\bar{Z} + 9 = 0 \), we can follow these steps:

### Step 1: Rewrite the line equation
The line equation can be expressed in terms of \( Z = x + iy \) and \( \bar{Z} = x - iy \):
\[ (3 + 4i)(x + iy) + (3 - 4i)(x - iy) + 9 = 0 \]

### Step 2: Expand the equation
Expanding the equation:
\[ (3x + 4iy + 3x - 4iy) + (9) + (3 - 4i)(x - iy) = 0 \]
This simplifies to:
\[ (3 + 3)x + (4i - 4i)y + 9 = 0 \]
Now, we need to expand \( (3 - 4i)(x - iy) \):
\[ = 3x - 3iy - 4ix + 4y = (3x + 4y) + i(-4x - 3y) \]

### Step 3: Combine real and imaginary parts
Combining the real and imaginary parts gives us:
\[ (3x + 3x + 4y + 9) + i(0) = 0 \]
This leads to the equation:
\[ 6x + 4y + 9 = 0 \]

### Step 4: Rearranging the line equation
Rearranging gives us:
\[ 6x + 4y = -9 \]

### Step 5: Identify coefficients
From the equation \( Ax + By + C = 0 \), we have:
- \( A = 6 \)
- \( B = 4 \)
- \( C = 9 \)

### Step 6: Use the distance formula
The distance \( d \) from a point \( (x_0, y_0) \) to a line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Substituting \( P(2, -3) \):
- \( x_0 = 2 \)
- \( y_0 = -3 \)

### Step 7: Calculate the distance
Substituting into the distance formula:
\[ d = \frac{|6(2) + 4(-3) + 9|}{\sqrt{6^2 + 4^2}} \]
Calculating the numerator:
\[ = |12 - 12 + 9| = |9| = 9 \]
Calculating the denominator:
\[ \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \]
Thus, the distance is:
\[ d = \frac{9}{2\sqrt{13}} = \frac{9\sqrt{13}}{26} \]

### Step 8: Final result
The length of the perpendicular from the point \( P(2 - 3i) \) to the line is:
\[ \frac{9\sqrt{13}}{26} \]