To find the equation of the line passing through point [tex]\( A(3,8) \)[/tex] and perpendicular to line segment [tex]\( \overline{BC} \)[/tex]:
1. Identify the coordinates of points [tex]\( B \)[/tex] and [tex]\( C \)[/tex]: [tex]\( B(7,5) \)[/tex] and [tex]\( C(2,3) \)[/tex].
2. Calculate the slope of [tex]\( \overline{BC} \)[/tex]:
[tex]\[
\text{slope}_{BC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 5}{2 - 7} = \frac{-2}{-5} = \frac{2}{5}
\][/tex]
3. The slope of the line perpendicular to [tex]\( \overline{BC} \)[/tex] is the negative reciprocal of the slope of [tex]\( \overline{BC} \)[/tex]:
[tex]\[
\text{slope}_{\text{perpendicular}} = -\frac{1}{\text{slope}_{BC}} = -\frac{1}{\frac{2}{5}} = -\frac{5}{2} = -2.5
\][/tex]
4. Use the point-slope form of the equation of the line that passes through point [tex]\( A(3, 8) \)[/tex]. The point-slope form equation is:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
where [tex]\((x_1, y_1) = (3, 8)\)[/tex] and [tex]\( m = -2.5 \)[/tex]:
[tex]\[
y - 8 = -2.5(x - 3)
\][/tex]
5. Simplify to the slope-intercept form [tex]\( y = mx + c \)[/tex]:
[tex]\[
y - 8 = -2.5x + 7.5
\][/tex]
[tex]\[
y = -2.5x + 7.5 + 8
\][/tex]
[tex]\[
y = -2.5x + 15.5
\][/tex]
So, the equation of the line passing through point [tex]\( A \)[/tex] and perpendicular to [tex]\( \overline{BC} \)[/tex] is:
[tex]\[
y = -2.5x + 15.5
\][/tex]
Therefore, filling in the boxes:
[tex]\( y = \boxed{-2.5} x + \boxed{15.5} \)[/tex]
