Find the distance of the intercept at point of x-axis and the line `5x+9y-45=0` from origin.

Options

A. 5
B. 9
C. `5//9`
D. `9//5`

Answer

To find the distance of the intercept at the point of the x-axis from the origin for the line given by the equation \(5x + 9y - 45 = 0\), we can follow these steps:

### Step 1: Find the x-intercept of the line
To find the x-intercept, we set \(y = 0\) in the equation of the line.

\[ 5x + 9(0) - 45 = 0 \]
This simplifies to:
\[ 5x - 45 = 0 \]
Now, solve for \(x\):
\[ 5x = 45 \implies x = \frac{45}{5} = 9 \]

### Step 2: Determine the coordinates of the intercept
The x-intercept we found is \(x = 9\) and since \(y = 0\) at the x-axis, the coordinates of the intercept point \(A\) are:
\[ A(9, 0) \]

### Step 3: Find the distance from the origin to the intercept point
The origin has coordinates \(O(0, 0)\). We can use the distance formula to find the distance \(d\) between the points \(A(9, 0)\) and \(O(0, 0)\):

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(9 - 0)^2 + (0 - 0)^2} \]
This simplifies to:
\[ d = \sqrt{9^2} = \sqrt{81} = 9 \]

### Final Answer
The distance of the intercept at the point of the x-axis from the origin is \(9\).

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