Answer :
Let's solve the problem step by step.
1. Finding the slope of the line [tex]\(\overleftrightarrow{A B}\)[/tex]:
- The coordinates of points [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are [tex]\((14, -1)\)[/tex] and [tex]\((2, 1)\)[/tex], respectively.
- The formula for the slope [tex]\( m_{AB} \)[/tex] of the line passing through these points is given by:
[tex]\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-1)}{2 - 14} = \frac{1 + 1}{2 - 14} = \frac{2}{-12} = -\frac{1}{6} \][/tex]
2. Finding the y-intercept of the line [tex]\(\overleftrightarrow{A B}\)[/tex]:
- The equation of the line in slope-intercept form is [tex]\( y = m_{AB} x + b_{AB} \)[/tex].
- Using the coordinates of point [tex]\(B\)[/tex], which is [tex]\((2, 1)\)[/tex]:
[tex]\[ 1 = -\frac{1}{6} \cdot 2 + b_{AB} \][/tex]
[tex]\[ 1 = -\frac{2}{6} + b_{AB} \][/tex]
[tex]\[ 1 = -\frac{1}{3} + b_{AB} \][/tex]
[tex]\[ b_{AB} = 1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3} \][/tex]
- Thus, the y-intercept of [tex]\(\overleftrightarrow{A B}\)[/tex] is [tex]\(\frac{4}{3}\)[/tex] or [tex]\(1.333333333\)[/tex].
3. Finding the equation of the line [tex]\(\overleftrightarrow{B C}\)[/tex]:
- Line [tex]\(\overleftrightarrow{B C}\)[/tex] is perpendicular to [tex]\(\overleftrightarrow{A B}\)[/tex], hence its slope [tex]\( m_{BC} \)[/tex] is the negative reciprocal of [tex]\( m_{AB} \)[/tex]:
[tex]\[ m_{BC} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{1}{6}} = 6 \][/tex]
- Using the point-slope form of the equation of the line and the coordinates of point [tex]\(B\)[/tex] (2, 1):
[tex]\[ y - 1 = 6(x - 2) \][/tex]
[tex]\[ y - 1 = 6x - 12 \][/tex]
[tex]\[ y = 6x - 12 + 1 \][/tex]
[tex]\[ y = 6x - 11 \][/tex]
- Therefore, the equation of the line [tex]\(\overleftrightarrow{B C}\)[/tex] is [tex]\( y = 6x - 11 \)[/tex].
4. Finding the x-coordinate of point [tex]\(C\)[/tex] when the y-coordinate of [tex]\(C\)[/tex] is 13:
[tex]\[ 13 = 6x - 11 \][/tex]
[tex]\[ 13 + 11 = 6x \][/tex]
[tex]\[ 24 = 6x \][/tex]
[tex]\[ x = \frac{24}{6} = 4 \][/tex]
- Thus, the x-coordinate of point [tex]\(C\)[/tex] is 4.
To summarize:
- The y-intercept of [tex]\(\overleftrightarrow{A B}\)[/tex] is [tex]\(1.333333333\)[/tex].
- The equation of [tex]\(\overleftrightarrow{B C}\)[/tex] is [tex]\( y = 6x - 11 \)[/tex].
- If the y-coordinate of point [tex]\(C\)[/tex] is 13, its x-coordinate is [tex]\(4\)[/tex].
So the correct answers for the boxes are:
1. [tex]\(\boxed{1.333333333}\)[/tex]
2. [tex]\(\boxed{6}\)[/tex]
3. [tex]\(\boxed{-11}\)[/tex]
4. [tex]\(\boxed{4}\)[/tex]
1. Finding the slope of the line [tex]\(\overleftrightarrow{A B}\)[/tex]:
- The coordinates of points [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are [tex]\((14, -1)\)[/tex] and [tex]\((2, 1)\)[/tex], respectively.
- The formula for the slope [tex]\( m_{AB} \)[/tex] of the line passing through these points is given by:
[tex]\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-1)}{2 - 14} = \frac{1 + 1}{2 - 14} = \frac{2}{-12} = -\frac{1}{6} \][/tex]
2. Finding the y-intercept of the line [tex]\(\overleftrightarrow{A B}\)[/tex]:
- The equation of the line in slope-intercept form is [tex]\( y = m_{AB} x + b_{AB} \)[/tex].
- Using the coordinates of point [tex]\(B\)[/tex], which is [tex]\((2, 1)\)[/tex]:
[tex]\[ 1 = -\frac{1}{6} \cdot 2 + b_{AB} \][/tex]
[tex]\[ 1 = -\frac{2}{6} + b_{AB} \][/tex]
[tex]\[ 1 = -\frac{1}{3} + b_{AB} \][/tex]
[tex]\[ b_{AB} = 1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3} \][/tex]
- Thus, the y-intercept of [tex]\(\overleftrightarrow{A B}\)[/tex] is [tex]\(\frac{4}{3}\)[/tex] or [tex]\(1.333333333\)[/tex].
3. Finding the equation of the line [tex]\(\overleftrightarrow{B C}\)[/tex]:
- Line [tex]\(\overleftrightarrow{B C}\)[/tex] is perpendicular to [tex]\(\overleftrightarrow{A B}\)[/tex], hence its slope [tex]\( m_{BC} \)[/tex] is the negative reciprocal of [tex]\( m_{AB} \)[/tex]:
[tex]\[ m_{BC} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{1}{6}} = 6 \][/tex]
- Using the point-slope form of the equation of the line and the coordinates of point [tex]\(B\)[/tex] (2, 1):
[tex]\[ y - 1 = 6(x - 2) \][/tex]
[tex]\[ y - 1 = 6x - 12 \][/tex]
[tex]\[ y = 6x - 12 + 1 \][/tex]
[tex]\[ y = 6x - 11 \][/tex]
- Therefore, the equation of the line [tex]\(\overleftrightarrow{B C}\)[/tex] is [tex]\( y = 6x - 11 \)[/tex].
4. Finding the x-coordinate of point [tex]\(C\)[/tex] when the y-coordinate of [tex]\(C\)[/tex] is 13:
[tex]\[ 13 = 6x - 11 \][/tex]
[tex]\[ 13 + 11 = 6x \][/tex]
[tex]\[ 24 = 6x \][/tex]
[tex]\[ x = \frac{24}{6} = 4 \][/tex]
- Thus, the x-coordinate of point [tex]\(C\)[/tex] is 4.
To summarize:
- The y-intercept of [tex]\(\overleftrightarrow{A B}\)[/tex] is [tex]\(1.333333333\)[/tex].
- The equation of [tex]\(\overleftrightarrow{B C}\)[/tex] is [tex]\( y = 6x - 11 \)[/tex].
- If the y-coordinate of point [tex]\(C\)[/tex] is 13, its x-coordinate is [tex]\(4\)[/tex].
So the correct answers for the boxes are:
1. [tex]\(\boxed{1.333333333}\)[/tex]
2. [tex]\(\boxed{6}\)[/tex]
3. [tex]\(\boxed{-11}\)[/tex]
4. [tex]\(\boxed{4}\)[/tex]
