Find the distance of a line `3x+4y+30=0` from origin.

Answer

To find the distance of the line \(3x + 4y + 30 = 0\) from the origin, we can use the formula for the distance \(d\) from a point \((x_0, y_0)\) to a line given by the equation \(Ax + By + C = 0\):

\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

### Step 1: Identify the coefficients
From the line equation \(3x + 4y + 30 = 0\), we can identify:
- \(A = 3\)
- \(B = 4\)
- \(C = 30\)

### Step 2: Substitute the coordinates of the origin
The coordinates of the origin are \((x_0, y_0) = (0, 0)\). We substitute these values into the distance formula:

\[ d = \frac{|3(0) + 4(0) + 30|}{\sqrt{3^2 + 4^2}} \]

### Step 3: Simplify the numerator
Calculating the numerator:

\[ d = \frac{|0 + 0 + 30|}{\sqrt{3^2 + 4^2}} = \frac{|30|}{\sqrt{3^2 + 4^2}} = \frac{30}{\sqrt{9 + 16}} = \frac{30}{\sqrt{25}} \]

### Step 4: Calculate the denominator
Now, calculate the denominator:

\[ \sqrt{25} = 5 \]

### Step 5: Final calculation
Now substitute back into the equation:

\[ d = \frac{30}{5} = 6 \]

Thus, the distance of the line \(3x + 4y + 30 = 0\) from the origin is \(6\) units.

### Summary of Solution:
The distance from the origin to the line \(3x + 4y + 30 = 0\) is \(6\) units.