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Sagot :
To determine the [tex]$y$[/tex]-intercept of the line segment [tex]$\overleftrightarrow{AB}$[/tex] and the equation of line [tex]$\overleftrightarrow{BC}$[/tex], let's go through the process step-by-step.
1. Find the slope of [tex]$\overleftrightarrow{A B}$[/tex]:
The slope of a line passing through points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Substituting the coordinates of points [tex]\(A(14, -1)\)[/tex] and [tex]\(B(2, 1)\)[/tex]:
[tex]\[ m_{AB} = \frac{1 - (-1)}{2 - 14} = \frac{1 + 1}{2 - 14} = \frac{2}{-12} = -\frac{1}{6} \][/tex]
2. Determine the [tex]$y$[/tex]-intercept of [tex]$\overleftrightarrow{A B}$[/tex]:
The equation of a line in slope-intercept form is [tex]\( y = mx + b \)[/tex]. Using the point-slope form [tex]\( y = mx + c \)[/tex] and substituting point [tex]\(B(2, 1)\)[/tex] and [tex]\( m_{AB} = -\frac{1}{6}\)[/tex]:
[tex]\[ 1 = -\frac{1}{6}(2) + c \\ 1 = -\frac{1}{3} + c \\ c = 1 + \frac{1}{3} \\ c = \frac{3}{3} + \frac{1}{3} \\ c = \frac{4}{3} \][/tex]
Thus, the [tex]$y$[/tex]-intercept of [tex]$\overleftrightarrow{A B}$[/tex] is [tex]\( \boxed{\frac{4}{3}} \)[/tex]
3. Find the slope of [tex]$\overleftrightarrow{B C}$[/tex]:
Since [tex]$\overleftrightarrow{AB}$[/tex] and [tex]$\overleftrightarrow{BC}$[/tex] form a right angle at point [tex]\(B\)[/tex], their slopes are negative reciprocals of each other:
[tex]\[ m_{BC} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{1}{6}} = 6 \][/tex]
4. Write the equation of [tex]$\overleftrightarrow{BC}$[/tex]:
Using the slope [tex]\( m_{BC} = 6 \)[/tex] and point [tex]\(B(2, 1)\)[/tex]:
[tex]\[ y - 1 = 6(x - 2) \\ y - 1 = 6x - 12 \\ y = 6x - 11 \][/tex]
The equation of [tex]$\overleftrightarrow{BC}$[/tex] is [tex]\( \boxed{y = 6x - 11} \)[/tex]
5. Find the [tex]$x$[/tex]-coordinate of point [tex]\(C\)[/tex] if the y-coordinate is 13:
Using the equation of [tex]\(\overleftrightarrow{BC}\)[/tex]:
[tex]\[ 13 = 6x - 11 \\ 13 + 11 = 6x \\ 24 = 6x \\ x = \frac{24}{6} \\ x = 4 \][/tex]
Therefore, the [tex]$x$[/tex]-coordinate of point [tex]\(C\)[/tex] is [tex]\( \boxed{4} \)[/tex]
1. Find the slope of [tex]$\overleftrightarrow{A B}$[/tex]:
The slope of a line passing through points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Substituting the coordinates of points [tex]\(A(14, -1)\)[/tex] and [tex]\(B(2, 1)\)[/tex]:
[tex]\[ m_{AB} = \frac{1 - (-1)}{2 - 14} = \frac{1 + 1}{2 - 14} = \frac{2}{-12} = -\frac{1}{6} \][/tex]
2. Determine the [tex]$y$[/tex]-intercept of [tex]$\overleftrightarrow{A B}$[/tex]:
The equation of a line in slope-intercept form is [tex]\( y = mx + b \)[/tex]. Using the point-slope form [tex]\( y = mx + c \)[/tex] and substituting point [tex]\(B(2, 1)\)[/tex] and [tex]\( m_{AB} = -\frac{1}{6}\)[/tex]:
[tex]\[ 1 = -\frac{1}{6}(2) + c \\ 1 = -\frac{1}{3} + c \\ c = 1 + \frac{1}{3} \\ c = \frac{3}{3} + \frac{1}{3} \\ c = \frac{4}{3} \][/tex]
Thus, the [tex]$y$[/tex]-intercept of [tex]$\overleftrightarrow{A B}$[/tex] is [tex]\( \boxed{\frac{4}{3}} \)[/tex]
3. Find the slope of [tex]$\overleftrightarrow{B C}$[/tex]:
Since [tex]$\overleftrightarrow{AB}$[/tex] and [tex]$\overleftrightarrow{BC}$[/tex] form a right angle at point [tex]\(B\)[/tex], their slopes are negative reciprocals of each other:
[tex]\[ m_{BC} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{1}{6}} = 6 \][/tex]
4. Write the equation of [tex]$\overleftrightarrow{BC}$[/tex]:
Using the slope [tex]\( m_{BC} = 6 \)[/tex] and point [tex]\(B(2, 1)\)[/tex]:
[tex]\[ y - 1 = 6(x - 2) \\ y - 1 = 6x - 12 \\ y = 6x - 11 \][/tex]
The equation of [tex]$\overleftrightarrow{BC}$[/tex] is [tex]\( \boxed{y = 6x - 11} \)[/tex]
5. Find the [tex]$x$[/tex]-coordinate of point [tex]\(C\)[/tex] if the y-coordinate is 13:
Using the equation of [tex]\(\overleftrightarrow{BC}\)[/tex]:
[tex]\[ 13 = 6x - 11 \\ 13 + 11 = 6x \\ 24 = 6x \\ x = \frac{24}{6} \\ x = 4 \][/tex]
Therefore, the [tex]$x$[/tex]-coordinate of point [tex]\(C\)[/tex] is [tex]\( \boxed{4} \)[/tex]
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