To find the equation of a line passing through point [tex]\( A(3, 8) \)[/tex] and perpendicular to the line segment [tex]\(\overline{BC}\)[/tex] where [tex]\( B(7, 5) \)[/tex] and [tex]\( C(2, 3) \)[/tex]:
1. Calculate the slope of [tex]\(\overline{BC}\)[/tex]:
[tex]\[
\text{Slope of } \overline{BC} = \frac{C_y - B_y}{C_x - B_x} = \frac{3 - 5}{2 - 7} = \frac{-2}{-5} = 0.4
\][/tex]
2. Determine the slope of the line perpendicular to [tex]\(\overline{BC}\)[/tex]:
[tex]\[
\text{Slope of the perpendicular line} = -\frac{1}{0.4} = -2.5
\][/tex]
3. Use the point-slope form of the equation of a line:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
where [tex]\( (x_1, y_1) = (3, 8) \)[/tex] and [tex]\( m = -2.5 \)[/tex].
4. Substitute the values into the point-slope form:
[tex]\[
y - 8 = -2.5(x - 3)
\][/tex]
5. Simplify to get the slope-intercept form ( [tex]\( y = mx + b \)[/tex] ):
[tex]\[
y - 8 = -2.5x + 7.5
\][/tex]
[tex]\[
y = -2.5x + 7.5 + 8
\][/tex]
[tex]\[
y = -2.5x + 15.5
\][/tex]
Thus, 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]
