Let's solve the problem step by step:
1. Finding the slope of line [tex]\( \overleftrightarrow{A B} \)[/tex]:
The coordinates of [tex]\( A \)[/tex] are [tex]\( A(14, -1) \)[/tex], and the coordinates of [tex]\( B \ ) are \( B(2, 1) \)[/tex].
The slope [tex]\( m \)[/tex] of the line passing through these points is given by the formula:
[tex]\[
\text{slope}_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-1)}{2 - 14} = \frac{2}{-12} = -\frac{1}{6}
\][/tex]
2. Finding the y-intercept of [tex]\( \overleftrightarrow{A B} \)[/tex]:
The equation of a line in slope-intercept form is [tex]\( y = mx + c \)[/tex]. We have the slope [tex]\( m = -\frac{1}{6} \)[/tex].
We use the coordinates of point [tex]\( B \)[/tex] (which are [tex]\( (2, 1) \)[/tex]) to find the y-intercept [tex]\( c \)[/tex]:
[tex]\[
y = -\frac{1}{6}x + c
\][/tex]
Substituting [tex]\( x = 2 \)[/tex] and [tex]\( y = 1 \)[/tex]:
[tex]\[
1 = -\frac{1}{6}(2) + c \implies 1 = -\frac{1}{3} + c \implies c = 1 + \frac{1}{3} = \frac{4}{3}
\][/tex]
So, the y-intercept of [tex]\( \overleftrightarrow{A B} \)[/tex] is [tex]\( \frac{4}{3} \)[/tex].
3. Finding the equation of line [tex]\( \overleftrightarrow{B C} \)[/tex]:
Since [tex]\( \overleftrightarrow{A B} \)[/tex] and [tex]\( \overleftrightarrow{B C} \)[/tex] form a right angle at point [tex]\( B \)[/tex], the slope of [tex]\( \overleftrightarrow{B C} \)[/tex] will be the negative reciprocal of the slope of [tex]\( \overleftrightarrow{A B} \)[/tex]:
[tex]\[
\text{slope}_{BC} = -\frac{1}{\text{slope}_{AB}} = -\frac{1}{-\frac{1}{6}} = 6
\][/tex]
Using the slope-intercept formula [tex]\( y = mx + c \)[/tex], point [tex]\( B(2, 1) \)[/tex], and knowing the slope is 6:
[tex]\[
y = 6x + c
\][/tex]
Substituting [tex]\( x = 2 \)[/tex] and [tex]\( y = 1 \)[/tex]:
[tex]\[
1 = 6(2) + c \implies 1 = 12 + c \implies c = 1 - 12 = -11
\][/tex]
Therefore, the equation of [tex]\( \overleftrightarrow{B C} \)[/tex] is:
[tex]\[
y = 6x - 11
\][/tex]
4. Finding the x-coordinate of point [tex]\( C \)[/tex] where the y-coordinate is 13:
The equation of line [tex]\( \overleftrightarrow{B C} \)[/tex] is [tex]\( y = 6x - 11 \)[/tex]. We set [tex]\( y = 13 \)[/tex]:
[tex]\[
13 = 6x - 11
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
13 + 11 = 6x \implies 24 = 6x \implies x = \frac{24}{6} = 4
\][/tex]
Thus, we have:
- The y-intercept of [tex]\( \overleftrightarrow{A B} \)[/tex] is [tex]\( \frac{4}{3} \)[/tex].
- The equation of [tex]\( \overleftrightarrow{B C} \)[/tex] is [tex]\( y = 6x - 11 \)[/tex].
- The x-coordinate of point [tex]\( C \)[/tex] (where the y-coordinate is 13) is 4.
So, the filled-in boxes should be:
- y-intercept of [tex]\( \overleftrightarrow{A B} \)[/tex]: [tex]\( \frac{4}{3} \)[/tex]
- Equation of [tex]\( \overleftrightarrow{B C} \)[/tex]: [tex]\( y = 6x - 11 \)[/tex]
- x-coordinate of [tex]\( C \)[/tex]: 4
