Find the equation of the line that has x-intercept = -3 and is perpendicular to `3x + 5y = 1`.

Answer

To find the equation of the line with an x-intercept of -3 and that is perpendicular to the line given by the equation \(3x + 5y = 1\), we can follow these steps:

### Step 1: Identify the x-intercept
The x-intercept is given as -3. This means the line passes through the point \((-3, 0)\).

### Step 2: Determine the slope of the given line
The equation of the given line is \(3x + 5y = 1\). We can rewrite this in slope-intercept form \(y = mx + b\) to find its slope.

1. Rearranging the equation:
\[ 5y = -3x + 1 \]
\[ y = -\frac{3}{5}x + \frac{1}{5} \]

The slope (\(m_1\)) of the given line is \(-\frac{3}{5}\).

### Step 3: Find the slope of the required line
Since the required line is perpendicular to the given line, we can use the property that the product of the slopes of two perpendicular lines is -1.

Let \(m_2\) be the slope of the required line. Then:
\[ m_1 \cdot m_2 = -1 \]
Substituting \(m_1 = -\frac{3}{5}\):
\[ -\frac{3}{5} \cdot m_2 = -1 \]
To find \(m_2\), we can rearrange:
\[ m_2 = \frac{5}{3} \]

### Step 4: Use the point-slope form to find the equation of the line
Now that we have the slope \(m_2 = \frac{5}{3}\) and a point \((-3, 0)\), we can use the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \(y_1 = 0\), \(m = \frac{5}{3}\), and \(x_1 = -3\):
\[ y - 0 = \frac{5}{3}(x - (-3)) \]
This simplifies to:
\[ y = \frac{5}{3}(x + 3) \]

### Step 5: Rearranging to standard form
Expanding the equation:
\[ y = \frac{5}{3}x + \frac{15}{3} \]
\[ y = \frac{5}{3}x + 5 \]
To convert this to standard form \(Ax + By + C = 0\), we can multiply through by 3 to eliminate the fraction:
\[ 3y = 5x + 15 \]
Rearranging gives:
\[ 5x - 3y + 15 = 0 \]

### Final Answer
The equation of the required line is:
\[ 5x - 3y + 15 = 0 \]