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[tex]$\overleftrightarrow{A B}$[/tex] and [tex]$\overleftrightarrow{B C}$[/tex] form a right angle at their point of intersection, [tex]$B$[/tex].

If the coordinates of [tex]$A$[/tex] and [tex]$B$[/tex] are [tex]$(14, -1)$[/tex] and [tex]$(2, 1)$[/tex], respectively, the [tex]$y$[/tex]-intercept of [tex]$\overleftrightarrow{A B}$[/tex] is [tex]$\square$[/tex].

The equation of [tex]$\overleftrightarrow{B C}$[/tex] is [tex]$y = \square x + \square$[/tex].

If the [tex]$y$[/tex]-coordinate of point [tex]$C$[/tex] is 13, its [tex]$x$[/tex]-coordinate is [tex]$\square$[/tex].

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To solve the problem, we need to find specific values and equations based on the given points and conditions. Here’s the step-by-step solution:

1. Calculate the slope of line [tex]\(\overleftrightarrow{AB}\)[/tex]:
- Point [tex]\(A\)[/tex] is [tex]\((14, -1)\)[/tex].
- Point [tex]\(B\)[/tex] is [tex]\((2, 1)\)[/tex].
- The slope [tex]\(m_{AB}\)[/tex] is given by:

[tex]\[ m_{AB} = \frac{y_B - y_A}{x_B - x_A} = \frac{1 - (-1)}{2 - 14} = \frac{2}{-12} = -\frac{1}{6} \][/tex]

2. Calculate the y-intercept of [tex]\(\overleftrightarrow{AB}\)[/tex]:
- Use the point-slope form [tex]\(y = mx + c\)[/tex] where [tex]\(m = -\frac{1}{6}\)[/tex] and point [tex]\(B\)[/tex].

[tex]\[ 1 = -\frac{1}{6} \cdot 2 + c \implies 1 = -\frac{1}{3} + c \implies c = 1 + \frac{1}{3} = \frac{4}{3} \][/tex]

So, the y-intercept of [tex]\(\overleftrightarrow{AB}\)[/tex] is [tex]\(\mathbf{\frac{4}{3}}\)[/tex].

3. Since [tex]\(\overleftrightarrow{AB}\)[/tex] and [tex]\(\overleftrightarrow{BC}\)[/tex] form a right angle at point [tex]\(B\)[/tex], the slope of [tex]\(\overleftrightarrow{BC}\)[/tex] is the negative reciprocal of the slope of [tex]\(\overleftrightarrow{AB}\)[/tex]:
- Slope [tex]\(m_{BC} = -\frac{1}{(-1/6)} = 6\)[/tex].

4. Find the equation of [tex]\(\overleftrightarrow{BC}\)[/tex]:
- Using slope [tex]\(m_{BC} = 6\)[/tex] and point [tex]\(B (2, 1)\)[/tex],

[tex]\[ y = 6x + c \][/tex]

[tex]\[ 1 = 6 \cdot 2 + c \implies 1 = 12 + c \implies c = 1 - 12 = -11 \][/tex]

So, the equation of [tex]\(\overleftrightarrow{BC}\)[/tex] is [tex]\(y = 6x - 11\)[/tex].

5. Calculate the x-coordinate of point [tex]\(C\)[/tex] when [tex]\(y_C = 13\)[/tex]:
- Using equation [tex]\(y = 6x - 11\)[/tex] with [tex]\(y_C = 13\)[/tex],

[tex]\[ 13 = 6x - 11 \][/tex]

[tex]\[ 6x = 13 + 11 = 24 \implies x = \frac{24}{6} = 4 \][/tex]

Therefore, the answers are:
- The y-intercept of [tex]\(\overleftrightarrow{A B}\)[/tex] is [tex]\(\mathbf{\frac{4}{3}}\)[/tex].
- The equation of [tex]\(\overleftrightarrow{B C}\)[/tex] is [tex]\(y = \mathbf{6x - 11}\)[/tex].
- If the [tex]\(y\)[/tex]-coordinate of point [tex]\(C\)[/tex] is [tex]\(13\)[/tex], its [tex]\(x\)[/tex]-coordinate is [tex]\(\mathbf{4}\)[/tex].
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