Find the distance between the line `12x-5y+15-0` and the point `(4,3)`.

Answer

To find the distance between the line \(12x - 5y + 15 = 0\) and 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 coefficients from the line equation**:
The line equation is given as \(12x - 5y + 15 = 0\). Here, we can identify:
- \(A = 12\)
- \(B = -5\)
- \(C = 15\)

2. **Substitute the point coordinates**:
The point given is \((4, 3)\), so we have:
- \(x_1 = 4\)
- \(y_1 = 3\)

3. **Plug values into the distance formula**:
Substitute \(A\), \(B\), \(C\), \(x_1\), and \(y_1\) into the distance formula:
\[ d = \frac{|12(4) + (-5)(3) + 15|}{\sqrt{12^2 + (-5)^2}} \]

4. **Calculate the numerator**:
Calculate \(12(4)\), \(-5(3)\), and then add \(15\):
\[ 12(4) = 48, \quad -5(3) = -15 \]
Therefore,
\[ 48 - 15 + 15 = 48 \]
Thus, the numerator becomes:
\[ |48| = 48 \]

5. **Calculate the denominator**:
Calculate \(A^2 + B^2\):
\[ A^2 = 12^2 = 144, \quad B^2 = (-5)^2 = 25 \]
Therefore,
\[ \sqrt{A^2 + B^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]

6. **Final calculation of distance**:
Now, substitute back into the formula:
\[ d = \frac{48}{13} \]

### Final Answer:
The distance between the line \(12x - 5y + 15 = 0\) and the point \((4, 3)\) is:
\[ d = \frac{48}{13} \text{ units} \]