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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]


Sagot :

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].