To find the equation of the line passing through point \( C \) and perpendicular to line segment \( \overline{AB} \), we start by determining the properties of line \( \overline{AB} \).
Given points \( A(2, 9) \) and \( B(8, 4) \):
1. Calculate the slope of \( \overline{AB} \):
[tex]\[
\text{slope of } \overline{AB} = \frac{B_y - A_y}{B_x - A_x} = \frac{4 - 9}{8 - 2} = \frac{-5}{6} = -\frac{5}{6}
\][/tex]
2. The slope of the line perpendicular to \( \overline{AB} \) is the negative reciprocal of the slope of \( \overline{AB} \):
[tex]\[
\text{slope of the perpendicular line} = -\left(-\frac{5}{6}\right)^{-1} = \frac{6}{5} = 1.2
\][/tex]
3. The equation of a line in slope-intercept form is \( y = mx + b \). We need to find the y-intercept \( b \) of the line that passes through point \( C \) with coordinates \( (-3, -2) \) and slope \( 1.2 \):
[tex]\[
y = 1.2x + b
\][/tex]
Substitute \( C(-3, -2) \) into the equation to solve for \( b \):
[tex]\[
-2 = 1.2(-3) + b
\][/tex]
[tex]\[
-2 = -3.6 + b
\][/tex]
[tex]\[
b = 1.6
\][/tex]
Therefore, the equation of the line passing through point \( C(-3, -2) \) and perpendicular to \( \overline{AB} \) is:
[tex]\[
y = 1.2x + 1.6
\][/tex]
So, completing the equation \( y = \square x + \square \):
[tex]\[
y = 1.2 \;x + 1.6
\][/tex]
