Answer :
To solve this problem, we must find the necessary characteristics of the lines and points based on the given information. Here are the steps:
### 1. Calculate the Slope of Line [tex]\( \overleftrightarrow{A B} \)[/tex]
Given points [tex]\( A(14, -1) \)[/tex] and [tex]\( B(2, 1) \)[/tex], we find the slope of line [tex]\( AB \)[/tex]:
[tex]\[ \text{slope}_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-1)}{2 - 14} = \frac{2}{-12} = -0.1667 \][/tex]
### 2. Determine the [tex]\( y \)[/tex]-Intercept of Line [tex]\( \overleftrightarrow{A B} \)[/tex]
Using the point-slope form of a line equation:
[tex]\[ y - y_1 = \text{slope} \cdot (x - x_1) \][/tex]
Choosing point [tex]\( B(2, 1) \)[/tex] for convenience:
[tex]\[ y - 1 = -0.1667(x - 2) \][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[ y = -0.1667x + 0.3333 + 1 \implies y = -0.1667x + 1.3333 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept ([tex]\( c \)[/tex]) of line [tex]\( \overleftrightarrow{A B} \)[/tex] is:
[tex]\[ c = 1.3333 \][/tex]
### 3. Calculate the Slope of Line [tex]\( \overleftrightarrow{B C} \)[/tex]
Since [tex]\( \overleftrightarrow{A B} \)[/tex] and [tex]\( \overleftrightarrow{B C} \)[/tex] form a right angle at point [tex]\( B \)[/tex], line [tex]\( \overleftrightarrow{B C} \)[/tex] must be perpendicular to [tex]\( \overleftrightarrow{A B} \)[/tex]. The slope of a line perpendicular to another is the negative reciprocal of the original slope.
Given the slope of [tex]\( \overleftrightarrow{A B} \)[/tex] is [tex]\( -0.1667 \)[/tex]:
[tex]\[ \text{slope}_{BC} = -\frac{1}{-0.1667} = 6.0 \][/tex]
### 4. Confirm Coordinates of Point [tex]\( C \)[/tex] Given Its [tex]\( y \)[/tex]-Coordinate is 13
Given [tex]\( y_C = 13 \)[/tex] and that we know [tex]\( \text{slope}_{BC}=6.0 \)[/tex], use the slope relationship of [tex]\( BC \)[/tex] passing through [tex]\( B(2, 1) \)[/tex]:
[tex]\[ y - y_B = \text{slope}_{BC} \cdot (x - x_B) \][/tex]
Substituting [tex]\( y_B = 1 \)[/tex], [tex]\( x_B = 2 \)[/tex], and [tex]\( \text{slope}_{BC} = 6 \)[/tex]:
[tex]\[ 13 - 1 = 6.0 \cdot (x - 2) \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ 12 = 6x - 12 \implies 6x = 24 \implies x = 4 \][/tex]
### Solution Summary
The solutions for this problem are:
- Slope of [tex]\( \overleftrightarrow{A B} \)[/tex]: [tex]\(-0.1667\)[/tex]
- [tex]\( y \)[/tex]-Intercept of [tex]\( \overleftrightarrow{A B} \)[/tex]: [tex]\( 1.3333\)[/tex]
- Slope of [tex]\( \overleftrightarrow{B C} \)[/tex]: [tex]\( 6.0 \)[/tex]
- [tex]\( x \)[/tex]-Coordinate of [tex]\( C \)[/tex]: [tex]\( 4 \)[/tex]
- [tex]\( y \)[/tex]-Coordinate of [tex]\( C \)[/tex]: [tex]\( 13 \)[/tex]
These values are verified based on the calculations and relationships given.
### 1. Calculate the Slope of Line [tex]\( \overleftrightarrow{A B} \)[/tex]
Given points [tex]\( A(14, -1) \)[/tex] and [tex]\( B(2, 1) \)[/tex], we find the slope of line [tex]\( AB \)[/tex]:
[tex]\[ \text{slope}_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-1)}{2 - 14} = \frac{2}{-12} = -0.1667 \][/tex]
### 2. Determine the [tex]\( y \)[/tex]-Intercept of Line [tex]\( \overleftrightarrow{A B} \)[/tex]
Using the point-slope form of a line equation:
[tex]\[ y - y_1 = \text{slope} \cdot (x - x_1) \][/tex]
Choosing point [tex]\( B(2, 1) \)[/tex] for convenience:
[tex]\[ y - 1 = -0.1667(x - 2) \][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[ y = -0.1667x + 0.3333 + 1 \implies y = -0.1667x + 1.3333 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept ([tex]\( c \)[/tex]) of line [tex]\( \overleftrightarrow{A B} \)[/tex] is:
[tex]\[ c = 1.3333 \][/tex]
### 3. Calculate the Slope of Line [tex]\( \overleftrightarrow{B C} \)[/tex]
Since [tex]\( \overleftrightarrow{A B} \)[/tex] and [tex]\( \overleftrightarrow{B C} \)[/tex] form a right angle at point [tex]\( B \)[/tex], line [tex]\( \overleftrightarrow{B C} \)[/tex] must be perpendicular to [tex]\( \overleftrightarrow{A B} \)[/tex]. The slope of a line perpendicular to another is the negative reciprocal of the original slope.
Given the slope of [tex]\( \overleftrightarrow{A B} \)[/tex] is [tex]\( -0.1667 \)[/tex]:
[tex]\[ \text{slope}_{BC} = -\frac{1}{-0.1667} = 6.0 \][/tex]
### 4. Confirm Coordinates of Point [tex]\( C \)[/tex] Given Its [tex]\( y \)[/tex]-Coordinate is 13
Given [tex]\( y_C = 13 \)[/tex] and that we know [tex]\( \text{slope}_{BC}=6.0 \)[/tex], use the slope relationship of [tex]\( BC \)[/tex] passing through [tex]\( B(2, 1) \)[/tex]:
[tex]\[ y - y_B = \text{slope}_{BC} \cdot (x - x_B) \][/tex]
Substituting [tex]\( y_B = 1 \)[/tex], [tex]\( x_B = 2 \)[/tex], and [tex]\( \text{slope}_{BC} = 6 \)[/tex]:
[tex]\[ 13 - 1 = 6.0 \cdot (x - 2) \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ 12 = 6x - 12 \implies 6x = 24 \implies x = 4 \][/tex]
### Solution Summary
The solutions for this problem are:
- Slope of [tex]\( \overleftrightarrow{A B} \)[/tex]: [tex]\(-0.1667\)[/tex]
- [tex]\( y \)[/tex]-Intercept of [tex]\( \overleftrightarrow{A B} \)[/tex]: [tex]\( 1.3333\)[/tex]
- Slope of [tex]\( \overleftrightarrow{B C} \)[/tex]: [tex]\( 6.0 \)[/tex]
- [tex]\( x \)[/tex]-Coordinate of [tex]\( C \)[/tex]: [tex]\( 4 \)[/tex]
- [tex]\( y \)[/tex]-Coordinate of [tex]\( C \)[/tex]: [tex]\( 13 \)[/tex]
These values are verified based on the calculations and relationships given.
