Question

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 β 2y 1 = 0:

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Maude

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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 \(y = \frac{5}{2}x - 8\) .

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