Let's solve the problem step-by-step:
1. Determine the slope of line \( \overleftrightarrow{A B} \):
The coordinates are \( A = (14, -1) \) and \( B = (2, 1) \).
The formula for the slope \(m\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:
[tex]\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
Here,
[tex]\[
m = \frac{1 - (-1)}{2 - 14} = \frac{2}{-12} = -\frac{1}{6}
\][/tex]
So, the slope of \( \overleftrightarrow{A B} \) is \(-\frac{1}{6}\).
2. Find the y-intercept of line \( \overleftrightarrow{A B} \):
The equation of a line in slope-intercept form is \( y = mx + c \). We already have \( m = -\frac{1}{6} \).
To find \( c \) (the y-intercept), we use the point \( B \) which is \((2, 1)\):
[tex]\[
1 = -\frac{1}{6} \cdot 2 + c
\][/tex]
Simplifying, we get:
[tex]\[
1 = -\frac{1}{3} + c \implies c = 1 + \frac{1}{3} = \frac{4}{3}
\][/tex]
So, the y-intercept is \(\frac{4}{3}\) and the equation of \( \overleftrightarrow{A B} \) is:
[tex]\[
y = -\frac{1}{6}x + \frac{4}{3}
\][/tex]
3. Find the slope of line \( \overleftrightarrow{B C} \):
Since \( \overleftrightarrow{B C} \) is perpendicular to \( \overleftrightarrow{A B} \), the product of their slopes is \(-1\).
Let \( m_{BC} \) be the slope of \( \overleftrightarrow{B C} \). Then:
[tex]\[
m_{BC} \cdot \left( -\frac{1}{6} \right) = -1 \implies m_{BC} = 6
\][/tex]
Thus, the slope of line \( \overleftrightarrow{B C} \) is \( 6 \).
4. Determine the x-coordinate of point \( C \) given its y-coordinate is 13:
The equation of \( \overleftrightarrow{B C} \) can be written using the slope-intercept form and the point \( B = (2, 1) \):
[tex]\[
y - 1 = 6(x - 2)
\][/tex]
Given \( y = 13 \):
[tex]\[
13 - 1 = 6(x - 2)
\][/tex]
Simplifying this equation:
[tex]\[
12 = 6(x - 2) \implies 2 = x - 2 \implies x = 4
\][/tex]
So, the x-coordinate of point \( C \) is \( 4 \).
Now, let's fill in the boxes in the original problem statement accordingly:
If the coordinates of \( A \) and \( B \) are \( (14, -1) \) and \( (2, 1) \), respectively, the \( y \)-intercept of \( \overleftrightarrow{A B} \) is \( y = -\frac{1}{6}x + \frac{4}{3} \).
If the [tex]\( y \)[/tex]-coordinate of point [tex]\( C \)[/tex] is [tex]\( 13 \)[/tex], its [tex]\( x \)[/tex]-coordinate is [tex]\( 4 \)[/tex].
