To create an equation for a line passing through point A(3, 8) and perpendicular to the line passing through points B(7, 5) and C(2, 3), follow these steps:
1. Find the slope of line BC:
The slope of a line through two points [tex]\( (x_1, y_1) \)[/tex] and [tex]\( (x_2, y_2) \)[/tex] is given by:
[tex]\[
\text{slope}_{BC} = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
For points B(7, 5) and C(2, 3), the slope is:
[tex]\[
\text{slope}_{BC} = \frac{3 - 5}{2 - 7} = \frac{-2}{-5} = \frac{2}{5}
\][/tex]
2. Determine the slope of the perpendicular line:
The slope of a line perpendicular to another line is the negative reciprocal of the slope of the original line. Thus, if the slope of BC is [tex]\(\frac{2}{5}\)[/tex], the slope of the perpendicular line is:
[tex]\[
\text{slope}_{\perp} = -\frac{1}{\left(\frac{2}{5}\right)} = -\frac{5}{2}
\][/tex]
3. Use the point-slope form to write the equation:
The point-slope form of a line's equation is:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
where [tex]\( m \)[/tex] is the slope and [tex]\( (x_1, y_1) \)[/tex] is a point on the line. Here [tex]\( (x_1, y_1) = (3, 8) \)[/tex] and [tex]\( m = -\frac{5}{2} \)[/tex]:
Substitute [tex]\( m = -\frac{5}{2} \)[/tex], [tex]\( x_1 = 3 \)[/tex], and [tex]\( y_1 = 8 \)[/tex]:
[tex]\[
y - 8 = -\frac{5}{2}(x - 3)
\][/tex]
4. Simplify the equation:
Distribute the slope on the right side:
[tex]\[
y - 8 = -\frac{5}{2}x + \frac{15}{2}
\][/tex]
Add 8 to both sides to solve for [tex]\( y \)[/tex]:
[tex]\[
y = -\frac{5}{2}x + \frac{15}{2} + 8
\][/tex]
Convert 8 to a fraction with the same denominator:
[tex]\[
y = -\frac{5}{2}x + \frac{15}{2} + \frac{16}{2}
\][/tex]
Combine the constants:
[tex]\[
y = -\frac{5}{2}x + \frac{31}{2}
\][/tex]
So, the equation of the line passing through point A(3, 8) and perpendicular to BC is:
[tex]\[
y = -\frac{5}{2}x + \frac{31}{2}
\][/tex]
