Determine the general equation of the straight line that passes through the point P (2;-3) and is parallel to the straight line with the equation 5x – 2y 1 = 0:

To find the equation of a line parallel to \(5x - 2y + 1 = 0\) and passing through point \(P(2, -3)\), we need to determine the slope of the given line first. The equation \(5x - 2y + 1 = 0\) can be rewritten in the slope-intercept form \(y = mx + c\) by rearranging it: \[5x - 2y + 1 = 0\] \[2y = 5x + 1\] \[y = \frac{5}{2}x + \frac{1}{2}\] From this form, we see that the slope of the line is \(m = \frac{5}{2}\). A line parallel to this line will have the same slope. So, the slope (\(m\)) of the line we're looking for is also \(m = \frac{5}{2}\). Now that we have the slope and the point \(P(2, -3)\), we can use the point-slope form of the equation of a line: \[y - y_1 = m(x - x_1)\] Substitute $4\(x_1 = 2\), \(y_1 = -3\), and \(m = \frac{5}{2}\)$$: \[y - (-3) = \frac{5}{2}(x - 2)\] \[y + 3 = \frac{5}{2}x - 5\] \[y = \frac{5}{2}x - 5 - 3\] \[y = \frac{5}{2}x - 8\] Therefore, the general equation of the straight line passing through point \(P(2, -3)\) and parallel to the line \(5x - 2y + 1 = 0\) is \(y = \frac{5}{2}x - 8\).