Answer :
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]
