Obtain the equation of a straight line passing through (3, 15) whose slope = 3\(\frac{1}{5}\).

To find the equation of a straight line with a given slope and a point through which it passes, we use the point-slope form of the line equation:

\[ y - y_1 = m(x - x_1) \]

Here:

• \((x_1, y_1)\) is the given point, which is \((3, 15)\)
• \(m\) is the slope, which is \(3 \frac{1}{5}\)

First, convert the mixed fraction to an improper fraction. \(3 \frac{1}{5} = \frac{16}{5}\).

Plug the values into the point-slope formula:

\[ y - 15 = \frac{16}{5}(x - 3) \]

Now, multiply both sides by 5 to eliminate the denominator:

\[ 5(y - 15) = 16(x - 3) \]

Expand both sides:

\[ 5y - 75 = 16x - 48 \]

Now, move all terms to one side to set the equation to zero:

\[ 5y - 16x - 75 + 48 = 0 \]

Simplify \(-75 + 48\):

\[ 5y - 16x - 27 = 0 \]

This is the equation of the straight line in general form (i.e., all terms on one side set to zero).

The correct answer uses the specific slope and passes through the given point, resulting in \[ 5y - 16x - 27 = 0 \]

This form shows how the formula is constructed and why each step is necessary based on the slope and point provided.