Let's solve the problem step-by-step.
1. Find the equation of the line [tex]\( \overleftrightarrow{AB} \)[/tex]:
- The coordinates of points [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are given as [tex]\( A(14, -1) \)[/tex] and [tex]\( B(2, 1) \)[/tex].
- First, let's calculate the slope [tex]\( m \)[/tex] of the line [tex]\( \overleftrightarrow{AB} \)[/tex]:
[tex]\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-1)}{2 - 14} = \frac{2}{-12} = -\frac{1}{6}
\][/tex]
- Next, we find the y-intercept [tex]\( b \)[/tex] of the line using the point-slope form of the equation [tex]\( y = mx + b \)[/tex]. Plugging in the coordinates of point [tex]\( B(2, 1) \)[/tex]:
[tex]\[
1 = -\frac{1}{6} \cdot 2 + b \implies 1 = -\frac{1}{3} + b \implies b = 1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3} \approx 1.33
\][/tex]
- Therefore, the equation of the line [tex]\( \overleftrightarrow{AB} \)[/tex] in the form [tex]\( y = mx + b \)[/tex] is:
[tex]\[
y = -\frac{1}{6}x + \frac{4}{3}
\][/tex]
2. Determine the coordinates of point [tex]\( C \)[/tex]:
- The y-coordinate of point [tex]\( C \)[/tex] is given as [tex]\( y_C = 13 \)[/tex].
- Since [tex]\( \overleftrightarrow{BC} \)[/tex] is perpendicular to [tex]\( \overleftrightarrow{AB} \)[/tex], the slope of [tex]\( \overleftrightarrow{BC} \)[/tex] is the negative reciprocal of the slope of [tex]\( \overleftrightarrow{AB} \)[/tex]. Thus, the slope of [tex]\( \overleftrightarrow{BC} \)[/tex] is:
[tex]\[
m_{BC} = -\left(-1 \div \frac{1}{6}\right) = 6
\][/tex]
- To find [tex]\( x_C \)[/tex], we will use the point-slope form and convert it to the slope-intercept form of [tex]\( \overleftrightarrow{BC} \)[/tex] passing through point [tex]\( B(2, 1) \)[/tex]:
[tex]\[
y - 1 = 6(x - 2)
\][/tex]
Simplifying for [tex]\( y \)[/tex]:
[tex]\[
y = 6x - 12 + 1 \implies y = 6x - 11
\][/tex]
Now, solve for [tex]\( x \)[/tex] when [tex]\( y_C = 13 \)[/tex]:
[tex]\[
13 = 6x - 11 \implies 13 + 11 = 6x \implies 24 = 6x \implies x = 4
\][/tex]
Thus, summarizing the solutions:
1. The equation of the line [tex]\( \overleftrightarrow{AB} \)[/tex] is:
[tex]\[
y = -\frac{1}{6} x + \frac{4}{3}
\][/tex]
2. The x-coordinate of point [tex]\( C \)[/tex] for [tex]\( y_C = 13 \)[/tex] is:
[tex]\[
x = 4
\][/tex]
