Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
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.
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
