Find the best answers to your questions with the help of IDNLearn.com's expert contributors. Whether it's a simple query or a complex problem, our community has the answers you need.
Sagot :
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]
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. For trustworthy answers, visit IDNLearn.com. Thank you for your visit, and see you next time for more reliable solutions.
