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Suppose a straight line is neither vertical nor horizontal but cuts the \(y\) axis at a certain point. This point is called the \(y\) - intercept of the line.
The equation of the slope intercept form of the line is given by:
\(y = mx + c\)
where \(m\) is the slope of the line, and
\(c\) is the \(y\) - intercept of the line.
Example:
1. Find the equation of the straight line whose slope is \(3\), and \(y\) - intercept is \(1\).
Solution:
Given that \(m = 3\) and \(c = 1\).
Substituting the known values in the equation of the slope intercept form, we get:
\(y = mx + c\)
\(y = 3x + 1\)
\(3x - y + 1 =0\)
Therefore, the equation of the straight line is \(3x - y + 1 =0\).
2. Determine the slope and \(y\) - intercept of the equation \(6x - 2y + 3 = 0\).
Solution:
The given equation is \(6x - 2y + 3 = 0\).
Let us write the given equation in the form of \(y = mx + c\).
Thus, \(-2y = -6x - 3\)
\(y = \frac{-6}{-2}x - \frac{3}{(-2)}\)
\(y = 3x + \frac{3}{2}\)
Comparing the above equation with the slope intercept form \(y = mc + c\), we have:
\(m = 3\) and \(c = \frac{3}{2}\)
Therefore, the slope is \(3\), and the \(y\)- intercept is \(\frac{3}{2}\).
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