Answer :
To find the solutions, we need to determine the slope and y-intercept of the lines as well as the x-coordinate of point [tex]\( C \)[/tex].
1. 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].
2. Calculate the slope of line [tex]\( AB \)[/tex]:
[tex]\[ \text{slope}_{AB} = \frac{(1 - (-1))}{(2 - 14)} = \frac{2}{-12} = -\frac{1}{6} \][/tex]
3. Using the point-slope form of the line equation with point [tex]\( B(2, 1) \)[/tex]:
[tex]\[ y - 1 = -\frac{1}{6}(x - 2) \][/tex]
Simplify to form the equation [tex]\( y = mx + c \)[/tex]:
[tex]\[ y = -\frac{1}{6}x + \left(1 + \frac{1}{6} \cdot 2\right) = -\frac{1}{6}x + \frac{4}{3} \][/tex]
Thus, the y-intercept of the line [tex]\( \overleftrightarrow{A B} \)[/tex] is:
[tex]\[ \boxed{\frac{4}{3}} \][/tex]
4. Line [tex]\( BC \)[/tex] is perpendicular to line [tex]\( AB \)[/tex], and the slope of [tex]\( BC \)[/tex] is the negative reciprocal of the slope of [tex]\( AB \)[/tex]:
[tex]\[ \text{slope}_{BC} = -\frac{1}{(-1/6)} = 6 \][/tex]
5. Using the point-slope form of the line equation with point [tex]\( B(2, 1) \)[/tex]:
[tex]\[ y - 1 = 6(x - 2) \][/tex]
Simplify to form the equation [tex]\( y = mx + c \)[/tex]:
[tex]\[ y = 6x - 12 + 1 = 6x - 11 \][/tex]
Thus, the equation of the line [tex]\( \overleftrightarrow{B C} \)[/tex] is:
[tex]\[ y = \boxed{6}x + \boxed{-11} \][/tex]
6. Finally, the given y-coordinate of point [tex]\( C \)[/tex] is 13. Solve for the x-coordinate by substituting [tex]\( y = 13 \)[/tex] into the equation of line [tex]\( BC \)[/tex]:
[tex]\[ 13 = 6x - 11 \][/tex]
[tex]\[ 6x = 24 \][/tex]
[tex]\[ x = 4 \][/tex]
Thus, the x-coordinate of point [tex]\( C \)[/tex] is:
[tex]\[ \boxed{4} \][/tex]
So, we have:
- The y-intercept of [tex]\( \overleftrightarrow{AB} \)[/tex] is [tex]\(\boxed{\frac{4}{3}}\)[/tex].
- The equation of [tex]\( \overleftrightarrow{BC} \)[/tex] is [tex]\( y = \boxed{6}x + \boxed{-11} \)[/tex].
- The x-coordinate of point [tex]\( C \)[/tex] is [tex]\( \boxed{4} \)[/tex].
1. 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].
2. Calculate the slope of line [tex]\( AB \)[/tex]:
[tex]\[ \text{slope}_{AB} = \frac{(1 - (-1))}{(2 - 14)} = \frac{2}{-12} = -\frac{1}{6} \][/tex]
3. Using the point-slope form of the line equation with point [tex]\( B(2, 1) \)[/tex]:
[tex]\[ y - 1 = -\frac{1}{6}(x - 2) \][/tex]
Simplify to form the equation [tex]\( y = mx + c \)[/tex]:
[tex]\[ y = -\frac{1}{6}x + \left(1 + \frac{1}{6} \cdot 2\right) = -\frac{1}{6}x + \frac{4}{3} \][/tex]
Thus, the y-intercept of the line [tex]\( \overleftrightarrow{A B} \)[/tex] is:
[tex]\[ \boxed{\frac{4}{3}} \][/tex]
4. Line [tex]\( BC \)[/tex] is perpendicular to line [tex]\( AB \)[/tex], and the slope of [tex]\( BC \)[/tex] is the negative reciprocal of the slope of [tex]\( AB \)[/tex]:
[tex]\[ \text{slope}_{BC} = -\frac{1}{(-1/6)} = 6 \][/tex]
5. Using the point-slope form of the line equation with point [tex]\( B(2, 1) \)[/tex]:
[tex]\[ y - 1 = 6(x - 2) \][/tex]
Simplify to form the equation [tex]\( y = mx + c \)[/tex]:
[tex]\[ y = 6x - 12 + 1 = 6x - 11 \][/tex]
Thus, the equation of the line [tex]\( \overleftrightarrow{B C} \)[/tex] is:
[tex]\[ y = \boxed{6}x + \boxed{-11} \][/tex]
6. Finally, the given y-coordinate of point [tex]\( C \)[/tex] is 13. Solve for the x-coordinate by substituting [tex]\( y = 13 \)[/tex] into the equation of line [tex]\( BC \)[/tex]:
[tex]\[ 13 = 6x - 11 \][/tex]
[tex]\[ 6x = 24 \][/tex]
[tex]\[ x = 4 \][/tex]
Thus, the x-coordinate of point [tex]\( C \)[/tex] is:
[tex]\[ \boxed{4} \][/tex]
So, we have:
- The y-intercept of [tex]\( \overleftrightarrow{AB} \)[/tex] is [tex]\(\boxed{\frac{4}{3}}\)[/tex].
- The equation of [tex]\( \overleftrightarrow{BC} \)[/tex] is [tex]\( y = \boxed{6}x + \boxed{-11} \)[/tex].
- The x-coordinate of point [tex]\( C \)[/tex] is [tex]\( \boxed{4} \)[/tex].