Find the distance of a line `4x+3y+13=0` from a point (4,3)

Answer

To find the distance of the line \(4x + 3y + 13 = 0\) from the point \((4, 3)\), we can use the formula for the distance \(d\) from a point \((x_1, y_1)\) to a line given by the equation \(Ax + By + C = 0\):

\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

### Step-by-step Solution:

1. **Identify the coefficients from the line equation**:
The line equation is given as \(4x + 3y + 13 = 0\).
- Here, \(A = 4\), \(B = 3\), and \(C = 13\).

2. **Identify the coordinates of the point**:
The point is given as \((x_1, y_1) = (4, 3)\).

3. **Substitute the values into the distance formula**:
Plugging in the values into the distance formula:
\[ d = \frac{|4(4) + 3(3) + 13|}{\sqrt{4^2 + 3^2}} \]

4. **Calculate the numerator**:
- Calculate \(4(4) = 16\)
- Calculate \(3(3) = 9\)
- Now, sum these values with \(C\):
\[ 16 + 9 + 13 = 38 \]
So, the numerator is \(|38| = 38\).

5. **Calculate the denominator**:
- Calculate \(A^2 + B^2\):
\[ 4^2 + 3^2 = 16 + 9 = 25 \]
- Now, take the square root:
\[ \sqrt{25} = 5 \]

6. **Calculate the distance**:
Now substitute the values back into the formula:
\[ d = \frac{38}{5} \]

### Final Answer:
Thus, the distance from the point \((4, 3)\) to the line \(4x + 3y + 13 = 0\) is:
\[ d = \frac{38}{5} \]