Answer :
Let's solve this step-by-step.
1. Find the slope of line [tex]\( \overleftrightarrow{AB} \)[/tex]:
The coordinates of points [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are [tex]\( (14, -1) \)[/tex] and [tex]\( (2, 1) \)[/tex], respectively.
The slope formula is:
[tex]\[ \text{slope}_{AB} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Substituting the coordinates [tex]\( (x_1, y_1) = (14, -1) \)[/tex] and [tex]\( (x_2, y_2) = (2, 1) \)[/tex]:
[tex]\[ \text{slope}_{AB} = \frac{1 - (-1)}{2 - 14} = \frac{2}{-12} = -\frac{1}{6} \approx -0.16666666666666666 \][/tex]
2. Find the y-intercept of line [tex]\( \overleftrightarrow{AB} \)[/tex]:
The equation of a line in slope-intercept form is [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept.
Using point [tex]\( B \)[/tex] [tex]\( (2, 1) \)[/tex]:
[tex]\[ 1 = -\frac{1}{6}(2) + b \][/tex]
Solving for [tex]\( b \)[/tex]:
[tex]\[ 1 = -\frac{1}{3} + b \implies b = 1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3} \approx 1.3333333333333333 \][/tex]
3. Construct the equation of line [tex]\( \overleftrightarrow{AB} \)[/tex]:
With slope [tex]\( m = -\frac{1}{6} \)[/tex] and y-intercept [tex]\( b = \frac{4}{3} \)[/tex]:
[tex]\[ y = -\frac{1}{6}x + \frac{4}{3} \][/tex]
4. Find the slope of line [tex]\( \stackrel{\leftrightarrow}{BC} \)[/tex]:
Since [tex]\( \overleftrightarrow{AB} \)[/tex] and [tex]\( \stackrel{\leftrightarrow}{BC} \)[/tex] form a right angle at point [tex]\( B \)[/tex], their slopes are negative reciprocals of each other.
The slope of line [tex]\( \stackrel{\leftrightarrow}{BC} \)[/tex] is:
[tex]\[ \text{slope}_{BC} = -\frac{1}{\text{slope}_{AB}} = -\frac{1}{-\frac{1}{6}} = 6 \][/tex]
5. Find the y-intercept of line [tex]\( \stackrel{\leftrightarrow}{BC} \)[/tex]:
Using the slope [tex]\( 6 \)[/tex] and point [tex]\( B (2, 1) \)[/tex]:
The equation form is [tex]\( y = 6x + b \)[/tex].
[tex]\[ 1 = 6(2) + b \implies 1 = 12 + b \implies b = 1 - 12 = -11 \][/tex]
6. Construct the equation of line [tex]\( \stackrel{\leftrightarrow}{BC} \)[/tex]:
With slope [tex]\( 6 \)[/tex] and y-intercept [tex]\( -11 \)[/tex]:
[tex]\[ y = 6x - 11 \][/tex]
7. Determine the x-coordinate of point [tex]\( C \)[/tex]:
Given that the y-coordinate of [tex]\( C \)[/tex] is 13, substitute [tex]\( y = 13 \)[/tex] into the equation of line [tex]\( \stackrel{\leftrightarrow}{BC} \)[/tex]:
[tex]\[ 13 = 6x - 11 \implies 13 + 11 = 6x \implies 24 = 6x \implies x = \frac{24}{6} = 4 \][/tex]
Therefore:
- The y-intercept of [tex]\( \overleftrightarrow{AB} \)[/tex] is [tex]\( \frac{4}{3} \)[/tex], or approximately [tex]\( 1.3333333333333333 \)[/tex].
- The equation of line [tex]\( \overleftrightarrow{AB} \)[/tex] is [tex]\( y = -\frac{1}{6} x + \frac{4}{3} \)[/tex].
- The x-coordinate of point [tex]\( C \)[/tex] is [tex]\( 4 \)[/tex].
Filling in the blanks from the question:
[tex]\[ \overleftrightarrow{A B} \text{ and } \stackrel{\leftrightarrow}{B C} \text{ form a right angle at their point of intersection, } B. \][/tex]
[tex]\[ \text{If the coordinates of } A \text{ and } B \text{ are } (14,-1) \text{ and } (2,1) \text{, respectively, the } y\text{-intercept of } \overleftrightarrow{A B} \text{ is } \frac{4}{3} \text{ or approximately } 1.3333333333333333. \][/tex]
[tex]\[ \text{The equation of line } \overleftrightarrow{A B} \text{ is } y = -\frac{1}{6} x + \frac{4}{3}. \][/tex]
[tex]\[ \text{If the } y\text{-coordinate of point } C \text{ is } 13 \text{, its } x\text{-coordinate is } 4. \][/tex]
1. Find the slope of line [tex]\( \overleftrightarrow{AB} \)[/tex]:
The coordinates of points [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are [tex]\( (14, -1) \)[/tex] and [tex]\( (2, 1) \)[/tex], respectively.
The slope formula is:
[tex]\[ \text{slope}_{AB} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Substituting the coordinates [tex]\( (x_1, y_1) = (14, -1) \)[/tex] and [tex]\( (x_2, y_2) = (2, 1) \)[/tex]:
[tex]\[ \text{slope}_{AB} = \frac{1 - (-1)}{2 - 14} = \frac{2}{-12} = -\frac{1}{6} \approx -0.16666666666666666 \][/tex]
2. Find the y-intercept of line [tex]\( \overleftrightarrow{AB} \)[/tex]:
The equation of a line in slope-intercept form is [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept.
Using point [tex]\( B \)[/tex] [tex]\( (2, 1) \)[/tex]:
[tex]\[ 1 = -\frac{1}{6}(2) + b \][/tex]
Solving for [tex]\( b \)[/tex]:
[tex]\[ 1 = -\frac{1}{3} + b \implies b = 1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3} \approx 1.3333333333333333 \][/tex]
3. Construct the equation of line [tex]\( \overleftrightarrow{AB} \)[/tex]:
With slope [tex]\( m = -\frac{1}{6} \)[/tex] and y-intercept [tex]\( b = \frac{4}{3} \)[/tex]:
[tex]\[ y = -\frac{1}{6}x + \frac{4}{3} \][/tex]
4. Find the slope of line [tex]\( \stackrel{\leftrightarrow}{BC} \)[/tex]:
Since [tex]\( \overleftrightarrow{AB} \)[/tex] and [tex]\( \stackrel{\leftrightarrow}{BC} \)[/tex] form a right angle at point [tex]\( B \)[/tex], their slopes are negative reciprocals of each other.
The slope of line [tex]\( \stackrel{\leftrightarrow}{BC} \)[/tex] is:
[tex]\[ \text{slope}_{BC} = -\frac{1}{\text{slope}_{AB}} = -\frac{1}{-\frac{1}{6}} = 6 \][/tex]
5. Find the y-intercept of line [tex]\( \stackrel{\leftrightarrow}{BC} \)[/tex]:
Using the slope [tex]\( 6 \)[/tex] and point [tex]\( B (2, 1) \)[/tex]:
The equation form is [tex]\( y = 6x + b \)[/tex].
[tex]\[ 1 = 6(2) + b \implies 1 = 12 + b \implies b = 1 - 12 = -11 \][/tex]
6. Construct the equation of line [tex]\( \stackrel{\leftrightarrow}{BC} \)[/tex]:
With slope [tex]\( 6 \)[/tex] and y-intercept [tex]\( -11 \)[/tex]:
[tex]\[ y = 6x - 11 \][/tex]
7. Determine the x-coordinate of point [tex]\( C \)[/tex]:
Given that the y-coordinate of [tex]\( C \)[/tex] is 13, substitute [tex]\( y = 13 \)[/tex] into the equation of line [tex]\( \stackrel{\leftrightarrow}{BC} \)[/tex]:
[tex]\[ 13 = 6x - 11 \implies 13 + 11 = 6x \implies 24 = 6x \implies x = \frac{24}{6} = 4 \][/tex]
Therefore:
- The y-intercept of [tex]\( \overleftrightarrow{AB} \)[/tex] is [tex]\( \frac{4}{3} \)[/tex], or approximately [tex]\( 1.3333333333333333 \)[/tex].
- The equation of line [tex]\( \overleftrightarrow{AB} \)[/tex] is [tex]\( y = -\frac{1}{6} x + \frac{4}{3} \)[/tex].
- The x-coordinate of point [tex]\( C \)[/tex] is [tex]\( 4 \)[/tex].
Filling in the blanks from the question:
[tex]\[ \overleftrightarrow{A B} \text{ and } \stackrel{\leftrightarrow}{B C} \text{ form a right angle at their point of intersection, } B. \][/tex]
[tex]\[ \text{If the coordinates of } A \text{ and } B \text{ are } (14,-1) \text{ and } (2,1) \text{, respectively, the } y\text{-intercept of } \overleftrightarrow{A B} \text{ is } \frac{4}{3} \text{ or approximately } 1.3333333333333333. \][/tex]
[tex]\[ \text{The equation of line } \overleftrightarrow{A B} \text{ is } y = -\frac{1}{6} x + \frac{4}{3}. \][/tex]
[tex]\[ \text{If the } y\text{-coordinate of point } C \text{ is } 13 \text{, its } x\text{-coordinate is } 4. \][/tex]
