Answer :
Let's solve the given problem step-by-step:
1. Finding the y-intercept of 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 slope of line [tex]\( \overleftrightarrow{A B} \)[/tex] is calculated using the formula [tex]\(\frac{y_2 - y_1}{x_2 - x_1}\)[/tex].
[tex]\[ \text{slope}_{AB} = \frac{1 - (-1)}{2 - 14} = \frac{2}{-12} = -\frac{1}{6} \][/tex]
- Using the slope-intercept form [tex]\( y = mx + b \)[/tex], we know the slope [tex]\( m \)[/tex] is [tex]\(-\frac{1}{6}\)[/tex] and the point [tex]\( B \)[/tex] [tex]\((2,1)\)[/tex] lies on the line. We substitute these values to find the y-intercept [tex]\( b \)[/tex]:
[tex]\[ 1 = -\frac{1}{6} \cdot 2 + b \implies 1 = -\frac{1}{3} + b \implies b = 1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3} \][/tex]
- Therefore, the y-intercept of [tex]\( \overleftrightarrow{A B} \)[/tex] is [tex]\( \boxed{\frac{4}{3}} \)[/tex].
2. Equation of line [tex]\( \overleftrightarrow{B C} \)[/tex]:
- Since lines [tex]\( \overleftrightarrow{A B} \)[/tex] and [tex]\( \overleftrightarrow{B C} \)[/tex] are perpendicular, the slope of [tex]\( \overleftrightarrow{B C} \)[/tex] is the negative reciprocal of the slope of [tex]\( \overleftrightarrow{A B} \)[/tex].
[tex]\[ \text{slope}_{BC} = -\frac{1}{-1/6} = 6 \][/tex]
- The slope of line [tex]\( \overleftrightarrow{B C} \)[/tex] is [tex]\( 6 \)[/tex].
- Using the point-slope form [tex]\( y = mx + b \)[/tex] with point [tex]\( B \)[/tex](2, 1) to find the y-intercept [tex]\( b \)[/tex]:
[tex]\[ 1 = 6 \cdot 2 + b \implies 1 = 12 + b \implies b = 1 - 12 = -11 \][/tex]
- Therefore, the equation of [tex]\( \overleftrightarrow{B C} \)[/tex] is [tex]\( y = 6x - 11 \)[/tex]. The equation for [tex]\( \overleftrightarrow{B C} \)[/tex] is: [tex]\( y = \boxed{6} x + \boxed{-11} \)[/tex].
3. Finding the x-coordinate of point [tex]\( C \)[/tex] given its y-coordinate is 13:
- Using the equation [tex]\( y = 6x - 11 \)[/tex] and [tex]\( y = 13 \)[/tex],
[tex]\[ 13 = 6x - 11 \implies 13 + 11 = 6x \implies 24 = 6x \implies x = \frac{24}{6} = 4 \][/tex]
- Therefore, the x-coordinate of point [tex]\( C \)[/tex] is [tex]\( \boxed{4} \)[/tex].
In summary:
- The y-intercept of [tex]\( \overleftrightarrow{A B} \)[/tex] is [tex]\( \boxed{\frac{4}{3}} \)[/tex].
- The equation of [tex]\( \overleftrightarrow{B C} \)[/tex] is [tex]\( y = \boxed{6} x + \boxed{-11} \)[/tex].
- The x-coordinate of point [tex]\( C \)[/tex] is [tex]\( \boxed{4} \)[/tex].
1. Finding the y-intercept of 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 slope of line [tex]\( \overleftrightarrow{A B} \)[/tex] is calculated using the formula [tex]\(\frac{y_2 - y_1}{x_2 - x_1}\)[/tex].
[tex]\[ \text{slope}_{AB} = \frac{1 - (-1)}{2 - 14} = \frac{2}{-12} = -\frac{1}{6} \][/tex]
- Using the slope-intercept form [tex]\( y = mx + b \)[/tex], we know the slope [tex]\( m \)[/tex] is [tex]\(-\frac{1}{6}\)[/tex] and the point [tex]\( B \)[/tex] [tex]\((2,1)\)[/tex] lies on the line. We substitute these values to find the y-intercept [tex]\( b \)[/tex]:
[tex]\[ 1 = -\frac{1}{6} \cdot 2 + b \implies 1 = -\frac{1}{3} + b \implies b = 1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3} \][/tex]
- Therefore, the y-intercept of [tex]\( \overleftrightarrow{A B} \)[/tex] is [tex]\( \boxed{\frac{4}{3}} \)[/tex].
2. Equation of line [tex]\( \overleftrightarrow{B C} \)[/tex]:
- Since lines [tex]\( \overleftrightarrow{A B} \)[/tex] and [tex]\( \overleftrightarrow{B C} \)[/tex] are perpendicular, the slope of [tex]\( \overleftrightarrow{B C} \)[/tex] is the negative reciprocal of the slope of [tex]\( \overleftrightarrow{A B} \)[/tex].
[tex]\[ \text{slope}_{BC} = -\frac{1}{-1/6} = 6 \][/tex]
- The slope of line [tex]\( \overleftrightarrow{B C} \)[/tex] is [tex]\( 6 \)[/tex].
- Using the point-slope form [tex]\( y = mx + b \)[/tex] with point [tex]\( B \)[/tex](2, 1) to find the y-intercept [tex]\( b \)[/tex]:
[tex]\[ 1 = 6 \cdot 2 + b \implies 1 = 12 + b \implies b = 1 - 12 = -11 \][/tex]
- Therefore, the equation of [tex]\( \overleftrightarrow{B C} \)[/tex] is [tex]\( y = 6x - 11 \)[/tex]. The equation for [tex]\( \overleftrightarrow{B C} \)[/tex] is: [tex]\( y = \boxed{6} x + \boxed{-11} \)[/tex].
3. Finding the x-coordinate of point [tex]\( C \)[/tex] given its y-coordinate is 13:
- Using the equation [tex]\( y = 6x - 11 \)[/tex] and [tex]\( y = 13 \)[/tex],
[tex]\[ 13 = 6x - 11 \implies 13 + 11 = 6x \implies 24 = 6x \implies x = \frac{24}{6} = 4 \][/tex]
- Therefore, the x-coordinate of point [tex]\( C \)[/tex] is [tex]\( \boxed{4} \)[/tex].
In summary:
- The y-intercept of [tex]\( \overleftrightarrow{A B} \)[/tex] is [tex]\( \boxed{\frac{4}{3}} \)[/tex].
- The equation of [tex]\( \overleftrightarrow{B C} \)[/tex] is [tex]\( y = \boxed{6} x + \boxed{-11} \)[/tex].
- The x-coordinate of point [tex]\( C \)[/tex] is [tex]\( \boxed{4} \)[/tex].
