To solve the question, we will determine the equation of a line passing through point [tex]\( C(-3, -2) \)[/tex] and perpendicular to line segment [tex]\(\overline{AB}\)[/tex].
In order to construct this line, we need to find the slope of line [tex]\(\overline{AB}\)[/tex] and subsequently, calculate the slope of the line that is perpendicular to it.
1. Find the slope of [tex]\(\overline{AB}\)[/tex]:
The slope [tex]\( m \)[/tex] of a line through two points [tex]\( (x_1, y_1) \)[/tex] and [tex]\( (x_2, y_2) \)[/tex] is given by:
[tex]\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
For points [tex]\( A(2, 9) \)[/tex] and [tex]\( B(8, 4) \)[/tex]:
[tex]\[
\text{slope}_{AB} = \frac{4 - 9}{8 - 2} = \frac{-5}{6}
\][/tex]
2. Calculate the slope of the perpendicular line:
The slope of a line that is perpendicular to another line is the negative reciprocal of that line's slope. Therefore, if the slope of [tex]\(\overline{AB}\)[/tex] is [tex]\(\frac{-5}{6}\)[/tex], then the slope of the line perpendicular to [tex]\(\overline{AB}\)[/tex] is:
[tex]\[
\text{slope}_{\text{perpendicular}} = -\frac{1}{\left( \frac{-5}{6} \right)} = \frac{6}{5}
\][/tex]
3. Form the equation of the line:
Using the point-slope form of a line equation [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:
[tex]\[
y - (-2) = \left( \frac{6}{5} \right)(x - (-3))
\][/tex]
Simplifying inside the parentheses:
[tex]\[
y + 2 = \frac{6}{5}(x + 3)
\][/tex]
4. Express the equation in slope-intercept form:
Expand and simplify:
[tex]\[
y + 2 = \frac{6}{5}x + \frac{6}{5} \times 3
\][/tex]
[tex]\[
y + 2 = \frac{6}{5}x + \frac{18}{5}
\][/tex]
Subtract 2 from both sides to isolate [tex]\( y \)[/tex]:
[tex]\[
y = \frac{6}{5}x + \frac{18}{5} - 2
\][/tex]
Converting 2 into a fraction with denominator 5:
[tex]\[
y = \frac{6}{5}x + \frac{18}{5} - \frac{10}{5}
\][/tex]
[tex]\[
y = \frac{6}{5}x + \frac{8}{5}
\][/tex]
So, the equation of the line passing through point [tex]\( C \)[/tex] and perpendicular to [tex]\(\overline{AB}\)[/tex] is:
[tex]\[
y = \frac{6}{5}x + \frac{8}{5}
\][/tex]
Therefore, the boxes should be filled as follows:
[tex]\[
y = 1.2x + 1.6
\][/tex]
