A line with positive rational slope, passes through the point A(6,0) and is at a distance of 5 units from B (1,3). The slope of line is

Options

A. `(15)/(8)`
B. `(8)/(15)`
C. `(5)/(8)`
D. `(8)/(5)`

Explanation

To find the slope of the line that passes through the point A(6,0) and is at a distance of 5 units from the point B(1,3), we can follow these steps:

### Step 1: Write the equation of the line
Let the slope of the line be \( m \). The equation of the line passing through the point (6, 0) can be written in point-slope form:
\[ y - 0 = m(x - 6) \]
This simplifies to:
\[ y = mx - 6m \]

### Step 2: Convert the line equation to standard form
Rearranging the equation gives us:
\[ mx - y - 6m = 0 \]

### Step 3: Calculate the perpendicular distance from point B to the line
The formula for the perpendicular distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
In our case, \( A = m \), \( B = -1 \), \( C = -6m \), and the point \( (x_0, y_0) = (1, 3) \). The distance is given as 5 units, so we set up the equation:
\[ \frac{|m(1) - 1(3) - 6m|}{\sqrt{m^2 + (-1)^2}} = 5 \]

### Step 4: Substitute and simplify
This simplifies to:
\[ \frac{|m - 3 - 6m|}{\sqrt{m^2 + 1}} = 5 \]
This further simplifies to:
\[ \frac{| -5m - 3 |}{\sqrt{m^2 + 1}} = 5 \]

### Step 5: Remove the absolute value and square both sides
We can consider two cases for the absolute value:

**Case 1**:
\[ -5m - 3 = 5\sqrt{m^2 + 1} \]
Squaring both sides:
\[ (5m + 3)^2 = 25(m^2 + 1) \]
Expanding and simplifying gives:
\[ 25m^2 + 30m + 9 = 25m^2 + 25 \]
\[ 30m + 9 - 25 = 0 \Rightarrow 30m - 16 = 0 \Rightarrow m = \frac{16}{30} = \frac{8}{15} \]

**Case 2**:
\[ -5m - 3 = -5\sqrt{m^2 + 1} \]
Squaring both sides gives the same equation as above, leading to the same result.

### Step 6: Conclusion
Thus, the slope of the line is:
\[ m = \frac{8}{15} \]