The distance of the point (2, 3) from the line `4x-3y+26=0` is same as its distance from the line `3x-4y+p=0`. The value of p can be

Options

A. 5
B. 25
C. 31
D. `-31`

Answer

To find the value of \( p \) such that the distance from the point \( (2, 3) \) to the line \( 4x - 3y + 26 = 0 \) is the same as its distance from the line \( 3x - 4y + p = 0 \), we will follow these steps:

### Step 1: Calculate the distance from the point \( (2, 3) \) to the line \( 4x - 3y + 26 = 0 \).

The formula for the distance \( D \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:

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

For the line \( 4x - 3y + 26 = 0 \):
- \( A = 4 \)
- \( B = -3 \)
- \( C = 26 \)

Substituting \( (x_0, y_0) = (2, 3) \):

\[ D_1 = \frac{|4(2) - 3(3) + 26|}{\sqrt{4^2 + (-3)^2}} \]

Calculating the numerator:

\[ 4(2) = 8, \quad -3(3) = -9 \quad \Rightarrow \quad 8 - 9 + 26 = 25 \]

Calculating the denominator:

\[ \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

Thus,

\[ D_1 = \frac{|25|}{5} = 5 \]

### Step 2: Set up the equation for the distance from the point \( (2, 3) \) to the line \( 3x - 4y + p = 0 \).

Using the same distance formula for the line \( 3x - 4y + p = 0 \):
- \( A = 3 \)
- \( B = -4 \)
- \( C = p \)

Substituting \( (x_0, y_0) = (2, 3) \):

\[ D_2 = \frac{|3(2) - 4(3) + p|}{\sqrt{3^2 + (-4)^2}} \]

Calculating the numerator:

\[ 3(2) = 6, \quad -4(3) = -12 \quad \Rightarrow \quad 6 - 12 + p = p - 6 \]

Calculating the denominator:

\[ \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

Thus,

\[ D_2 = \frac{|p - 6|}{5} \]

### Step 3: Set the distances equal.

Since \( D_1 = D_2 \):

\[ 5 = \frac{|p - 6|}{5} \]

Multiplying both sides by 5:

\[ 25 = |p - 6| \]

### Step 4: Solve for \( p \).

This absolute value equation gives us two cases:

1. \( p - 6 = 25 \)
2. \( p - 6 = -25 \)

**Case 1:**

\[ p - 6 = 25 \quad \Rightarrow \quad p = 31 \]

**Case 2:**

\[ p - 6 = -25 \quad \Rightarrow \quad p = -19 \]

### Conclusion

The possible values of \( p \) are \( 31 \) and \( -19 \).