Let's tackle the given problems one by one.
### Problem 1: Finding the equation of a line parallel to [tex]\( y = 3x + 5 \)[/tex] passing through the point [tex]\( (5, 9) \)[/tex].
Firstly, recall the point-slope form of a line's equation:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\((x_1, y_1)\)[/tex] is a point on the line and [tex]\(m\)[/tex] is the slope.
Since we are looking for a line parallel to [tex]\(y = 3x + 5\)[/tex], it will have the same slope. The slope ([tex]\(m\)[/tex]) of the given line [tex]\(y = 3x + 5\)[/tex] is [tex]\(3\)[/tex].
We are also given the point [tex]\((5,9)\)[/tex] through which the new line passes. Let's use this point and the slope [tex]\(3\)[/tex] in the point-slope form:
[tex]\[ y - 9 = 3(x - 5) \][/tex]
So, the equation of the line parallel to [tex]\(y=3x+5\)[/tex] and passing through [tex]\((5,9)\)[/tex] is:
[tex]\[ y - 9 = 3(x - 5) \][/tex]
### Problem 2: Finding the equation of a line parallel to [tex]\( y = \frac{1}{2} x \)[/tex] with a [tex]\( y \)[/tex]-intercept of [tex]\((0, 4)\)[/tex] in slope-intercept form.
Firstly, recall the slope-intercept form of a line's equation:
[tex]\[ y = mx + b \][/tex]
where [tex]\(m\)[/tex] is the slope and [tex]\(b\)[/tex] is the [tex]\(y\)[/tex]-intercept.
The given line is [tex]\( y = \frac{1}{2}x \)[/tex]. This line has a slope ([tex]\(m\)[/tex]) of [tex]\(\frac{1}{2}\)[/tex].
We want a line parallel to this slope with a [tex]\( y \)[/tex]-intercept [tex]\( (0, 4) \)[/tex]. Thus, the intercept [tex]\( b \)[/tex] for our new line is [tex]\(4\)[/tex].
Now, using the slope [tex]\( \frac{1}{2} \)[/tex] and the intercept [tex]\( 4 \)[/tex], we can write the equation of the line in slope-intercept form:
[tex]\[ y = \frac{1}{2}x + 4 \][/tex]
So, the equation of the line parallel to [tex]\( y = \frac{1}{2} x \)[/tex] with a [tex]\( y \)[/tex]-intercept of [tex]\(0, 4\)[/tex] is:
[tex]\[ y = \frac{1}{2} x + 4 \][/tex]
