Get personalized answers to your specific questions with IDNLearn.com. Ask your questions and get detailed, reliable answers from our community of experienced experts.
Sagot :
To solve the problem step-by-step:
1. Determine the slope of line AB:
The slope ([tex]\( m \)[/tex]) between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
For points [tex]\( A(14, -1) \)[/tex] and [tex]\( B(2, 1) \)[/tex]:
[tex]\[ m = \frac{1 - (-1)}{2 - 14} = \frac{2}{-12} = -\frac{1}{6} \][/tex]
So, the slope of line AB is [tex]\(-\frac{1}{6}\)[/tex].
2. Determine the y-intercept of line AB:
The equation of a line in slope-intercept form is [tex]\( y = mx + c \)[/tex]. Plugging in the slope and using point [tex]\( B(2, 1) \)[/tex] to find the y-intercept ([tex]\( c \)[/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 line AB is [tex]\( \frac{4}{3} \)[/tex].
3. Determine the slope of line BC:
Since lines AB and BC form a right angle, the slopes of these lines are negative reciprocals of each other. Thus, the slope of line BC is:
[tex]\[ m_{\text{BC}} = -\frac{1}{-\frac{1}{6}} = 6 \][/tex]
4. Determine the equation of line BC:
Using the slope [tex]\( m_{\text{BC}} = 6 \)[/tex] and point [tex]\( B(2, 1) \)[/tex]:
[tex]\[ y = 6x + c \][/tex]
To find the y-intercept ([tex]\( c \)[/tex]):
[tex]\[ 1 = 6(2) + c \implies 1 = 12 + c \implies c = 1 - 12 = -11 \][/tex]
So, the equation of line BC is [tex]\( y = 6x - 11 \)[/tex].
5. Determine the x-coordinate of point C:
Given the y-coordinate of point C ([tex]\( y_C \)[/tex]) is 13, and using the equation of line BC:
[tex]\[ 13 = 6x_C - 11 \implies 13 + 11 = 6x_C \implies 24 = 6x_C \implies x_C = \frac{24}{6} = 4 \][/tex]
So, summarizing the results:
- The [tex]\( y \)[/tex]-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 [tex]\( x \)[/tex]-coordinate of point [tex]\( C \)[/tex] is [tex]\( 4 \)[/tex].
The completed answer is:
- [tex]\( \boxed{\frac{4}{3}} \)[/tex]
- [tex]\( \boxed{6} \)[/tex]
- [tex]\( \boxed{-11} \)[/tex]
- [tex]\( \boxed{4} \)[/tex]
1. Determine the slope of line AB:
The slope ([tex]\( m \)[/tex]) between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
For points [tex]\( A(14, -1) \)[/tex] and [tex]\( B(2, 1) \)[/tex]:
[tex]\[ m = \frac{1 - (-1)}{2 - 14} = \frac{2}{-12} = -\frac{1}{6} \][/tex]
So, the slope of line AB is [tex]\(-\frac{1}{6}\)[/tex].
2. Determine the y-intercept of line AB:
The equation of a line in slope-intercept form is [tex]\( y = mx + c \)[/tex]. Plugging in the slope and using point [tex]\( B(2, 1) \)[/tex] to find the y-intercept ([tex]\( c \)[/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 line AB is [tex]\( \frac{4}{3} \)[/tex].
3. Determine the slope of line BC:
Since lines AB and BC form a right angle, the slopes of these lines are negative reciprocals of each other. Thus, the slope of line BC is:
[tex]\[ m_{\text{BC}} = -\frac{1}{-\frac{1}{6}} = 6 \][/tex]
4. Determine the equation of line BC:
Using the slope [tex]\( m_{\text{BC}} = 6 \)[/tex] and point [tex]\( B(2, 1) \)[/tex]:
[tex]\[ y = 6x + c \][/tex]
To find the y-intercept ([tex]\( c \)[/tex]):
[tex]\[ 1 = 6(2) + c \implies 1 = 12 + c \implies c = 1 - 12 = -11 \][/tex]
So, the equation of line BC is [tex]\( y = 6x - 11 \)[/tex].
5. Determine the x-coordinate of point C:
Given the y-coordinate of point C ([tex]\( y_C \)[/tex]) is 13, and using the equation of line BC:
[tex]\[ 13 = 6x_C - 11 \implies 13 + 11 = 6x_C \implies 24 = 6x_C \implies x_C = \frac{24}{6} = 4 \][/tex]
So, summarizing the results:
- The [tex]\( y \)[/tex]-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 [tex]\( x \)[/tex]-coordinate of point [tex]\( C \)[/tex] is [tex]\( 4 \)[/tex].
The completed answer is:
- [tex]\( \boxed{\frac{4}{3}} \)[/tex]
- [tex]\( \boxed{6} \)[/tex]
- [tex]\( \boxed{-11} \)[/tex]
- [tex]\( \boxed{4} \)[/tex]
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. IDNLearn.com has the answers you need. Thank you for visiting, and we look forward to helping you again soon.