First, we need to find the slope of line segment [tex]\( \overline{A B} \)[/tex]. The formula for the slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[
\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
Using the points [tex]\( A(2, 9) \)[/tex] and [tex]\( B(8, 4) \)[/tex]:
[tex]\[
\text{slope of } \overline{A B} = \frac{4 - 9}{8 - 2} = \frac{-5}{6}
\][/tex]
Next, for a line perpendicular to [tex]\( \overline{A B} \)[/tex], the slope is the negative reciprocal of [tex]\(-\frac{5}{6}\)[/tex]. Thus:
[tex]\[
\text{slope of the perpendicular line} = \frac{6}{5}
\][/tex]
We then use the point-slope form of the equation for the line that passes through point [tex]\( C(-3, -2) \)[/tex] with this slope. The point-slope form is:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
Plugging in the values, [tex]\((x_1, y_1) = (-3, -2)\)[/tex] and [tex]\(m = \frac{6}{5}\)[/tex]:
[tex]\[
y - (-2) = \frac{6}{5}(x - (-3))
\][/tex]
This simplifies to:
[tex]\[
y + 2 = \frac{6}{5}(x + 3)
\][/tex]
To convert this to the slope-intercept form [tex]\( y = mx + b \)[/tex], we need to solve for [tex]\( y \)[/tex]:
[tex]\[
y + 2 = \frac{6}{5}x + \frac{6}{5} \cdot 3
\][/tex]
[tex]\[
y + 2 = \frac{6}{5}x + \frac{18}{5}
\][/tex]
Subtracting 2 from both sides to solve for [tex]\( y \)[/tex]:
[tex]\[
y = \frac{6}{5}x + \frac{18}{5} - 2
\][/tex]
[tex]\[
y = \frac{6}{5}x + \frac{18}{5} - \frac{10}{5}
\][/tex]
[tex]\[
y = \frac{6}{5}x + \frac{8}{5}
\][/tex]
Thus, the equation of the line is:
[tex]\[
y = 1.2x + 1.6
\][/tex]
So for the boxes:
[tex]\[
y = \boxed{1.2}x + \boxed{1.6}
\][/tex]
