Find the equation of the line that is parallel to [tex][tex]$y=\frac{1}{2} x-4$[/tex][/tex] and contains the point [tex](4, 5)[/tex].

[tex] y = \frac{1}{2} x + [\quad] [/tex]



Answer :

Certainly! To find the equation of a line that is parallel to a given line and passes through a specific point, we can follow these steps:

1. Identify the slope of the given line.

The given line is [tex]\( y = \frac{1}{2}x - 4 \)[/tex]. The slope of this line is [tex]\(\frac{1}{2}\)[/tex].

2. Determine the slope of the parallel line.

Lines that are parallel have the same slope, so the slope of the new line will also be [tex]\(\frac{1}{2}\)[/tex].

3. Use the point-slope form of the equation of a line.

The point-slope form of a line's equation is given by:
[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. We know the slope [tex]\( m = \frac{1}{2} \)[/tex] and the line passes through the point [tex]\( (4, 5) \)[/tex].

4. Substitute the known values into the point-slope form.

Plugging in the slope and the coordinates of the point [tex]\( (4, 5) \)[/tex]:
[tex]\[ y - 5 = \frac{1}{2}(x - 4) \][/tex]

5. Simplify the equation to slope-intercept form [tex]\( y = mx + b \)[/tex].

To do this, we need to distribute [tex]\(\frac{1}{2}\)[/tex] and then solve for [tex]\( y \)[/tex]:
[tex]\[ y - 5 = \frac{1}{2}x - 2 \][/tex]

Now, add 5 to both sides to isolate [tex]\( y \)[/tex]:
[tex]\[ y = \frac{1}{2}x - 2 + 5 \][/tex]

Combine like terms:
[tex]\[ y = \frac{1}{2}x + 3 \][/tex]

So, the equation of the line that is parallel to [tex]\(y = \frac{1}{2}x - 4\)[/tex] and passes through the point [tex]\( (4, 5) \)[/tex] is:
[tex]\[ y = \frac{1}{2}x + 3 \][/tex]

Therefore, the final equation you are looking for is:
[tex]\[ y = \frac{1}{2} x + 3 \][/tex]