Find the distancce of the point P(-2, 3) from the line AB which is `x-y=5`.

Answer

To find the distance of the point P(-2, 3) from the line given by the equation \( x - y = 5 \), we can follow these steps:

### Step 1: Rewrite the line equation in standard form
The given line equation is:
\[ x - y = 5 \]
We can rewrite it in the standard form \( Ax + By + C = 0 \):
\[ x - y - 5 = 0 \]
Here, we have:
- \( A = 1 \)
- \( B = -1 \)
- \( C = -5 \)

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

### Step 3: Substitute the values into the formula
Now we substitute \( A \), \( B \), \( C \), \( x_0 \), and \( y_0 \) into the distance formula:
\[ d = \frac{|1(-2) + (-1)(3) - 5|}{\sqrt{1^2 + (-1)^2}} \]

### Step 4: Calculate the numerator
Calculating the numerator:
\[ = | -2 - 3 - 5 | = |-10| = 10 \]

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

### Step 6: Calculate the distance
Now, substituting back into the distance formula:
\[ d = \frac{10}{\sqrt{2}} = 10 \cdot \frac{\sqrt{2}}{2} = 5\sqrt{2} \]

### Final Answer
Thus, the distance of the point P(-2, 3) from the line \( x - y = 5 \) is:
\[ \boxed{5\sqrt{2}} \]