Question

Determine the equation of one straight line which is parallel to the line 3x + 5y =1 and crosses the x-axis at -3

Answer 1

To determine the equation of a line that is parallel to the line \(3x + 5y = 1\) and crosses the x-axis at \(-3\), we can follow these steps:

1. **Find the slope of the given line**:
The equation of the line \(3x + 5y = 1\) can be rewritten in slope-intercept form \(y = mx + b\):
\[
5y = -3x + 1
\]
\[
y = -\frac{3}{5}x + \frac{1}{5}
\]
From this, we can see that the slope \(m\) of the line \(3x + 5y = 1\) is \(-\frac{3}{5}\).

2. **Use the slope for the parallel line**:
Since parallel lines have the same slope, the slope of the new line will also be \(-\frac{3}{5}\).

3. **Find the y-coordinate when x = -3**:
A line crosses the x-axis at \((-3, 0)\).

4. **Use the point-slope form of the line**:
The point-slope form of a line is given by:
\[
y - y_1 = m(x - x_1)
\]
Here, the point \((x_1, y_1)\) is \((-3, 0)\) and the slope \(m\) is \(-\frac{3}{5}\).
\[
y - 0 = -\frac{3}{5}(x - (-3))
\]
\[
y = -\frac{3}{5}(x + 3)
\]
\[
y = -\frac{3}{5}x - \frac{9}{5}
\]

5. **Convert to standard form (optional)**:
To convert the equation into standard form \(Ax + By = C\):
\[
\frac{3}{5}x + y = -\frac{9}{5}
\]
Multiply through by 5 to eliminate the fraction:
\[
3x + 5y = -9
\]

Thus, the equation of the line that is parallel to \(3x + 5y = 1\) and crosses the x-axis at \(-3\) is:
\[
3x + 5y = -9
\]