To create an equation for a line passing through point [tex]\( A(3,8) \)[/tex] and perpendicular to line segment [tex]\( \overline{BC} \)[/tex], we need to follow these steps:
1. Find the slope of line [tex]\( BC \)[/tex]:
[tex]\[
\text{slope of } BC = \frac{y_C - y_B}{x_C - x_B}
\][/tex]
Plugging in the coordinates [tex]\( B(7,5) \)[/tex] and [tex]\( C(2,3) \)[/tex]:
[tex]\[
\text{slope of } BC = \frac{3 - 5}{2 - 7} = \frac{-2}{-5} = \frac{2}{5}
\][/tex]
2. Find the slope of the line perpendicular to [tex]\( BC \)[/tex]:
The slope of the line perpendicular to another line is the negative reciprocal of the slope of the original line.
Since the slope of [tex]\( BC \)[/tex] is [tex]\( \frac{2}{5} \)[/tex]:
[tex]\[
\text{slope of the perpendicular line} = -\frac{1}{(\frac{2}{5})} = -\frac{5}{2}
\][/tex]
3. Use point-slope form to find the equation of the line passing through [tex]\( A(3,8) \)[/tex]:
The point-slope form of a line's equation is:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
Here, [tex]\( (x_1, y_1) = (3, 8) \)[/tex] and [tex]\( m = -2.5 \)[/tex]:
[tex]\[
y - 8 = -2.5(x - 3)
\][/tex]
4. Solve for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] to get the equation in slope-intercept form ( [tex]\( y = mx + b \)[/tex] ):
Distribute and solve the equation:
[tex]\[
y - 8 = -2.5x + 7.5
\][/tex]
Adding 8 to both sides:
[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]
So, the correct answer is:
[tex]\[ y = -2.5x + 15.5 \][/tex]
