In this problem, we need to find which equation represents the line [tex]\(\overleftrightarrow{ AC }\)[/tex] that is perpendicular to the line [tex]\(\overleftrightarrow{ DB }\)[/tex]. The given equation of [tex]\(\overleftrightarrow{ DB }\)[/tex] is:
[tex]\[
\frac{1}{2}x + 2y = 12
\][/tex]
First, let's convert this equation into slope-intercept form ([tex]\(y = mx + b\)[/tex]), where [tex]\(m\)[/tex] represents the slope of the line.
[tex]\[
\frac{1}{2}x + 2y = 12
\][/tex]
Isolate [tex]\(y\)[/tex]:
[tex]\[
2y = -\frac{1}{2}x + 12
\][/tex]
Divide both sides by 2:
[tex]\[
y = -\frac{1}{4}x + 6
\][/tex]
So, the slope of line [tex]\(\overleftrightarrow{ DB }\)[/tex] is [tex]\(-\frac{1}{4}\)[/tex].
Since lines [tex]\(\overleftrightarrow{ DB }\)[/tex] and [tex]\(\overleftrightarrow{ AC }\)[/tex] are perpendicular, the slope of [tex]\(\overleftrightarrow{ AC }\)[/tex] should be the negative reciprocal of [tex]\(-\frac{1}{4}\)[/tex]. To find the negative reciprocal, we flip the fraction and change the sign.
The negative reciprocal of [tex]\(-\frac{1}{4}\)[/tex] is:
[tex]\[
4
\][/tex]
Now we need to find which of the given options has a slope of 4:
1. [tex]\(-4x + y = -28\)[/tex]
Convert to slope-intercept form:
[tex]\[
y = 4x - 28
\][/tex]
The slope of this line is 4.
2. [tex]\(2x + y = 14\)[/tex]
Convert to slope-intercept form:
[tex]\[
y = -2x + 14
\][/tex]
The slope of this line is -2.
3. [tex]\(2x + 8y = 12\)[/tex]
Convert to slope-intercept form:
[tex]\[
8y = -2x + 12
\][/tex]
[tex]\[
y = -\frac{1}{4}x + \frac{3}{2}
\][/tex]
The slope of this line is -[tex]\(\frac{1}{4}\)[/tex].
4. [tex]\(4x - y = -28\)[/tex]
Convert to slope-intercept form:
[tex]\[
-y = -4x - 28
\][/tex]
[tex]\[
y = 4x + 28
\][/tex]
The slope of this line is 4.
From the above conversions, we see that both option 1 ([tex]\(-4x + y = -28\)[/tex]) and option 4 ([tex]\(4x - y = -28\)[/tex]) have a slope of 4, which matches the requirement for being perpendicular to the line [tex]\(\overleftrightarrow{ DB }\)[/tex].
Therefore, the correct equations of [tex]\(\overleftrightarrow{ AC }\)[/tex] are:
[tex]\[
\boxed{-4x + y = -28 \ \text{and} \ 4x - y = -28}
\][/tex]
