Find the distance between the st.line `4x+3y-5=0` and the point `(-2,1)`.

Answer

To find the distance between the straight line \(4x + 3y - 5 = 0\) and the point \((-2, 1)\), 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 given line equation is \(4x + 3y - 5 = 0\).
Here, we can identify:
- \(A = 4\)
- \(B = 3\)
- \(C = -5\)

2. **Identify the coordinates of the point**:
The point given is \((-2, 1)\).
Thus, we have:
- \(x_1 = -2\)
- \(y_1 = 1\)

3. **Substitute the values into the distance formula**:
Plugging in the values into the distance formula:
\[ d = \frac{|4(-2) + 3(1) - 5|}{\sqrt{4^2 + 3^2}} \]

4. **Calculate the numerator**:
First, calculate \(4(-2) + 3(1) - 5\):
\[ 4(-2) = -8 \]
\[ 3(1) = 3 \]
\[ -8 + 3 - 5 = -10 \]
Now take the absolute value:
\[ | -10 | = 10 \]

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

6. **Calculate the distance**:
Now substitute back into the formula:
\[ d = \frac{10}{5} = 2 \]

### Final Answer:
The distance between the line \(4x + 3y - 5 = 0\) and the point \((-2, 1)\) is \(2\).