The length iof perpendicular from `(3,1) `on line `4x+3y+20 =0,` is

Options

A. 6
B. 7
C. 5
D. 8

Answer

To find the length of the perpendicular from the point (3, 1) to the line given by the equation \(4x + 3y + 20 = 0\), we can use the formula for the distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\):

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

### Step-by-Step Solution:

1. **Identify the coefficients**: From the line equation \(4x + 3y + 20 = 0\), we can identify:
- \(A = 4\)
- \(B = 3\)
- \(C = 20\)

2. **Substitute the point coordinates**: The point given is \((x_1, y_1) = (3, 1)\).

3. **Plug values into the formula**:
\[ d = \frac{|4(3) + 3(1) + 20|}{\sqrt{4^2 + 3^2}} \]

4. **Calculate the numerator**:
\[ 4(3) = 12 \]
\[ 3(1) = 3 \]
\[ 12 + 3 + 20 = 35 \]
Thus, the numerator becomes:
\[ |35| = 35 \]

5. **Calculate the denominator**:
\[ \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

6. **Calculate the distance**:
\[ d = \frac{35}{5} = 7 \]

### Final Answer:
The length of the perpendicular from the point (3, 1) to the line \(4x + 3y + 20 = 0\) is **7**.

---