### Theory:

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 $$\frac{3}{2}$$.