First, we'll start by finding the slope of the given line defined by the equation \(3x = 5y + 3\). We can rearrange it to slope-intercept form \(y = mx + b\).
- Rearranging the equation: \[ 3x - 3 = 5y \] \[ 5y = 3x - 3 \] \[ y = \frac{3}{5}x - \frac{3}{5} \]
From the equation, we see that the slope \(m\) of the given line is \(\frac{3}{5}\).
Since we want a line that is parallel to this one, it will have the same slope, \(\frac{3}{5}\). Now we use the point-slope form of the equation of a line, which is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point the line passes through. Here, \((x_1, y_1) = (5, 2)\).
- Using the point-slope form: \[ y - 2 = \frac{3}{5}(x - 5) \]
Now let's simplify this into slope-intercept form: \[ y - 2 = \frac{3}{5}x - \frac{15}{5} \] \[ y - 2 = \frac{3}{5}x - 3 \] \[ y = \frac{3}{5}x - 1 \]
This is the equation in slope-intercept form: \[ y = \frac{3}{5}x - 1 \]
- Now, we need to convert this to standard form \(Ax + By = C\) with integer coefficients.
Starting from the slope-intercept form: \[ y = \frac{3}{5}x - 1 \] Multiply through by 5 to eliminate the fraction: \[ 5y = 3x - 5 \]
Rearranging gives: \[ -3x + 5y = -5 \]
To write it in standard form with positive coefficients, we can multiply the entire equation by -1: \[ 3x - 5y = 5 \]
So, the final answers are:
In slope-intercept form: \[ y = \frac{3}{5}x - 1 \]
In standard form: \[ 3x - 5y = 5 \]