To find the equation of the line passing through point [tex]\( C \)[/tex] and perpendicular to line segment [tex]\( \overline{AB} \)[/tex], we follow these steps:
1. Calculate the slope of line [tex]\( \overline{AB} \)[/tex]:
[tex]\[ \text{slope}_{AB} = \frac{B_y - A_y}{B_x - A_x} = \frac{4 - 9}{8 - 2} = \frac{-5}{6} \][/tex]
2. Calculate the slope of the line perpendicular to [tex]\( \overline{AB} \)[/tex]:
[tex]\[ \text{slope}_{\text{perpendicular}} = -\frac{1}{\text{slope}_{AB}} = -\frac{1}{-\frac{5}{6}} = \frac{6}{5} \][/tex]
Here, we see that the perpendicular slope [tex]\( \frac{6}{5} \)[/tex] approximates to [tex]\( 1.2 \)[/tex].
3. Find the y-intercept of the line passing through point [tex]\( C(-3, -2) \)[/tex]:
We use the point-slope form of the line equation [tex]\( y - y_1 = m(x - x1) \)[/tex], where [tex]\( (x_1, y_1) \)[/tex] is point [tex]\( C \)[/tex] and [tex]\( m \)[/tex] is the slope of the perpendicular line.
[tex]\[ y + 2 = \frac{6}{5} (x + 3) \][/tex]
4. To find the y-intercept, we simplify the equation to the slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[ y + 2 = 1.2(x + 3) \][/tex]
[tex]\[ y + 2 = 1.2x + 3.6 \][/tex]
[tex]\[ y = 1.2x + 3.6 - 2 \][/tex]
[tex]\[ y = 1.2x + 1.6 \][/tex]
So, the equation of the line passing through point [tex]\( C \)[/tex] and perpendicular to [tex]\( \overline{A B} \)[/tex] is:
[tex]\[ y = 1.2x + 1.6 \][/tex]
Therefore, the correct answer is:
[tex]\[
y = \boxed{1.2} x + \boxed{1.6}
\][/tex]
