Line t contains the points (1,1) and (-5,3). Plot a point other than point P with integral coordinates that is on a line perpendicular to line t and passes through point P.
Line t contains the points (1,1) and (-5,3). Plot a point other than point P with integral coordinates that is on a line perpendicular to line t and passes through point P.
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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![## Plotting a Point on a Perpendicular Line
**Task:**
Given Line \( t \) which contains the points \((1,1)\) and \((-5,3)\), plot a point other than point \( P \) with integral coordinates that lies on a line perpendicular to line \( t \) and passes through point \( P \).
### Instructions:
1. **Identify Line \( t \):**
- Line \( t \) is drawn through the points \((1,1)\) and \((-5,3)\).
2. **Graph Details:**
- The graph is a standard coordinate plane with x and y-axes, intersecting at the origin \((0,0)\).
- Point \( P \) is located on line \( t \).
3. **Determining the Slope of Line \( t \):**
- Calculate the slope (m) using the formula:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]
- Using points \((1,1)\) and \((-5,3)\),
\[
m = \frac{3 - 1}{-5 - 1} = \frac{2}{-6} = -\frac{1}{3}
\]
4. **Finding the Slope of the Perpendicular Line:**
- The slope of a line perpendicular to line \( t \) is the negative reciprocal of \(-\frac{1}{3}\).
\[
\text{slope of perpendicular line} = 3
\]
5. **Equation of the Perpendicular Line:**
- Use point-slope form to find the equation of the perpendicular line passing through point \( P \):
\[
y - y_1 = m(x - x_1)
\]
- Substitute point \( P \)’s coordinates into the equation (assuming \( P = (a,b) \)):
\[
y - b = 3(x - a)
\]
6. **Plotting an Integral Point on the Perpendicular Line:**
- Choose a point with integer coordinates that lies on this line.
- Ensure the point selected passes the equation criteria.
### Example Solution:
Let's assume \( P = (0,0) \) for simplicity; the line equation becomes](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc3f0117e-26e1-4209-8b43-253a6b092b43%2Fcbc83822-4113-46bc-90e4-abcd89278b24%2Fvn9zabv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Plotting a Point on a Perpendicular Line
**Task:**
Given Line \( t \) which contains the points \((1,1)\) and \((-5,3)\), plot a point other than point \( P \) with integral coordinates that lies on a line perpendicular to line \( t \) and passes through point \( P \).
### Instructions:
1. **Identify Line \( t \):**
- Line \( t \) is drawn through the points \((1,1)\) and \((-5,3)\).
2. **Graph Details:**
- The graph is a standard coordinate plane with x and y-axes, intersecting at the origin \((0,0)\).
- Point \( P \) is located on line \( t \).
3. **Determining the Slope of Line \( t \):**
- Calculate the slope (m) using the formula:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]
- Using points \((1,1)\) and \((-5,3)\),
\[
m = \frac{3 - 1}{-5 - 1} = \frac{2}{-6} = -\frac{1}{3}
\]
4. **Finding the Slope of the Perpendicular Line:**
- The slope of a line perpendicular to line \( t \) is the negative reciprocal of \(-\frac{1}{3}\).
\[
\text{slope of perpendicular line} = 3
\]
5. **Equation of the Perpendicular Line:**
- Use point-slope form to find the equation of the perpendicular line passing through point \( P \):
\[
y - y_1 = m(x - x_1)
\]
- Substitute point \( P \)’s coordinates into the equation (assuming \( P = (a,b) \)):
\[
y - b = 3(x - a)
\]
6. **Plotting an Integral Point on the Perpendicular Line:**
- Choose a point with integer coordinates that lies on this line.
- Ensure the point selected passes the equation criteria.
### Example Solution:
Let's assume \( P = (0,0) \) for simplicity; the line equation becomes
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