First, calculate the slope, [tex]\( m \)[/tex], of line [tex]\(\overleftrightarrow{A B}\)[/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, use point [tex]\( B \)[/tex] (2, 1) to find the y-intercept [tex]\( c \)[/tex] of [tex]\(\overleftrightarrow{A B}\)[/tex] with the slope-intercept form of a line:
[tex]\[ y = mx + c \][/tex]
Substituting in the coordinates of [tex]\( B \)[/tex] and the slope:
[tex]\[ 1 = -\frac{1}{6}(2) + c \][/tex]
[tex]\[ 1 = -\frac{1}{3} + c \][/tex]
[tex]\[ c = 1 + \frac{1}{3} = \frac{4}{3} \][/tex]
So, the y-intercept of [tex]\(\overleftrightarrow{A B}\)[/tex] is [tex]\( \frac{4}{3} \)[/tex].
For the line [tex]\(\overleftrightarrow{B C}\)[/tex], which is perpendicular to [tex]\(\overleftrightarrow{A B}\)[/tex], the slope must be the negative reciprocal of [tex]\(-\frac{1}{6}\)[/tex]:
[tex]\[ \text{slope of } \overleftrightarrow{B C} = 6 \][/tex]
Using the point [tex]\( B \)[/tex] again to find the y-intercept of [tex]\(\overleftrightarrow{B C}\)[/tex]:
[tex]\[ y = mx + c \][/tex]
[tex]\[ 1 = 6(2) + c \][/tex]
[tex]\[ 1 = 12 + c \][/tex]
[tex]\[ c = 1 - 12 = -11 \][/tex]
So, the equation of [tex]\(\overleftrightarrow{B C}\)[/tex] is:
[tex]\[ y = 6x - 11 \][/tex]
If the y-coordinate of point [tex]\( C \)[/tex] is 13, substitute [tex]\( y = 13 \)[/tex] in the equation of [tex]\(\overleftrightarrow{B C}\)[/tex] to find the x-coordinate [tex]\( x \)[/tex]:
[tex]\[ 13 = 6x - 11 \][/tex]
[tex]\[ 13 + 11 = 6x \][/tex]
[tex]\[ 24 = 6x \][/tex]
[tex]\[ x = \frac{24}{6} = 4 \][/tex]
Therefore, the y-intercept of [tex]\(\overleftrightarrow{A B}\)[/tex] is [tex]\(\boxed{\frac{4}{3}}\)[/tex], the equation of [tex]\(\overleftrightarrow{B C}\)[/tex] is [tex]\( y = \boxed{6} x + \boxed{-11} \)[/tex], and the x-coordinate of point [tex]\( C \)[/tex] is [tex]\( \boxed{4} \)[/tex].
