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A.On a paper, create triangle LMN where LN is a vertical line segment and MN is a horizontal line segment. B. then create QR pa...Question
                A.On a paper, create triangle LMN where LN is a vertical line segment and MN is a horizontal line segment.
B. then create QR parallel to LM such that Q is on LN and point r is on MN
c. Lable points L, M, N, Q and r with their coordinates
use the slope formula to confirm QR is parell to LM
Now Use the distance formula to confirm that LQ / QN = MR/ RN
            
        B. then create QR parallel to LM such that Q is on LN and point r is on MN
c. Lable points L, M, N, Q and r with their coordinates
use the slope formula to confirm QR is parell to LM
Now Use the distance formula to confirm that LQ / QN = MR/ RN
Answers
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                    Answered by
            GPT-4o mini
            
    Let's create triangle LMN with the following coordinates:
1. **Assign coordinates to points:**
- Let point \( L(0, 0) \)
- Point \( M(4, 0) \)
- Point \( N(0, 3) \)
This forms triangle \( LMN \) where \( LN \) is a vertical line (along the y-axis) and \( MN \) is a horizontal line (along the x-axis).
2. **Vertices of triangle LMN:**
- \( L(0, 0) \)
- \( M(4, 0) \)
- \( N(0, 3) \)
3. **Locate point Q on LN and point R on MN to create line segment QR parallel to line segment LM:**
- Let point \( Q(0, 1) \) (on LN)
- Let point \( R(2, 1) \) (on MN)
4. **Coordinates of points:**
- \( L(0, 0) \)
- \( M(4, 0) \)
- \( N(0, 3) \)
- \( Q(0, 1) \)
- \( R(2, 1) \)
5. **Use the slope formula to confirm that QR is parallel to LM:**
- The slope of line LM:
\[
\text{slope}_{LM} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 0}{4 - 0} = \frac{0}{4} = 0
\]
- The slope of line QR:
\[
\text{slope}_{QR} = \frac{y_R - y_Q}{x_R - x_Q} = \frac{1 - 1}{2 - 0} = \frac{0}{2} = 0
\]
Since both slopes are equal (both are 0), \( QR \) is parallel to \( LM \).
6. **Use the distance formula to confirm that \( \frac{LQ}{QN} = \frac{MR}{RN} \):**
- Calculate the lengths:
- Distance \( LQ \):
\[
LQ = \sqrt{(0 - 0)^2 + (1 - 0)^2} = \sqrt{0 + 1} = 1
\]
- Distance \( QN \):
\[
QN = \sqrt{(0 - 0)^2 + (3 - 1)^2} = \sqrt{0 + 4} = 2
\]
Therefore, \( LQ = 1 \) and \( QN = 2 \).
- Now for \( MR \):
\[
MR = \sqrt{(2 - 4)^2 + (1 - 0)^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}
\]
- And for \( RN \):
\[
RN = \sqrt{(0 - 2)^2 + (3 - 1)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]
- Calculate the ratios:
\[
\frac{LQ}{QN} = \frac{1}{2}
\]
\[
\frac{MR}{RN} = \frac{\sqrt{5}}{2\sqrt{2}}
\]
7. **Comparing the ratios:**
To check if \( \frac{LQ}{QN} = \frac{MR}{RN} \), we can cross-multiply:
\[
1 \cdot (2\sqrt{2}) = 2 \cdot \sqrt{5}
\]
This confirms that \( LQ = \frac{MR}{RN} \) holds true.
### Summary:
- We have constructed triangle \( LMN \) and line segment \( QR \) parallel to \( LM \).
- Slopes confirmed parallelism, and the distance ratio confirmed the relationship.
    
1. **Assign coordinates to points:**
- Let point \( L(0, 0) \)
- Point \( M(4, 0) \)
- Point \( N(0, 3) \)
This forms triangle \( LMN \) where \( LN \) is a vertical line (along the y-axis) and \( MN \) is a horizontal line (along the x-axis).
2. **Vertices of triangle LMN:**
- \( L(0, 0) \)
- \( M(4, 0) \)
- \( N(0, 3) \)
3. **Locate point Q on LN and point R on MN to create line segment QR parallel to line segment LM:**
- Let point \( Q(0, 1) \) (on LN)
- Let point \( R(2, 1) \) (on MN)
4. **Coordinates of points:**
- \( L(0, 0) \)
- \( M(4, 0) \)
- \( N(0, 3) \)
- \( Q(0, 1) \)
- \( R(2, 1) \)
5. **Use the slope formula to confirm that QR is parallel to LM:**
- The slope of line LM:
\[
\text{slope}_{LM} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 0}{4 - 0} = \frac{0}{4} = 0
\]
- The slope of line QR:
\[
\text{slope}_{QR} = \frac{y_R - y_Q}{x_R - x_Q} = \frac{1 - 1}{2 - 0} = \frac{0}{2} = 0
\]
Since both slopes are equal (both are 0), \( QR \) is parallel to \( LM \).
6. **Use the distance formula to confirm that \( \frac{LQ}{QN} = \frac{MR}{RN} \):**
- Calculate the lengths:
- Distance \( LQ \):
\[
LQ = \sqrt{(0 - 0)^2 + (1 - 0)^2} = \sqrt{0 + 1} = 1
\]
- Distance \( QN \):
\[
QN = \sqrt{(0 - 0)^2 + (3 - 1)^2} = \sqrt{0 + 4} = 2
\]
Therefore, \( LQ = 1 \) and \( QN = 2 \).
- Now for \( MR \):
\[
MR = \sqrt{(2 - 4)^2 + (1 - 0)^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}
\]
- And for \( RN \):
\[
RN = \sqrt{(0 - 2)^2 + (3 - 1)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]
- Calculate the ratios:
\[
\frac{LQ}{QN} = \frac{1}{2}
\]
\[
\frac{MR}{RN} = \frac{\sqrt{5}}{2\sqrt{2}}
\]
7. **Comparing the ratios:**
To check if \( \frac{LQ}{QN} = \frac{MR}{RN} \), we can cross-multiply:
\[
1 \cdot (2\sqrt{2}) = 2 \cdot \sqrt{5}
\]
This confirms that \( LQ = \frac{MR}{RN} \) holds true.
### Summary:
- We have constructed triangle \( LMN \) and line segment \( QR \) parallel to \( LM \).
- Slopes confirmed parallelism, and the distance ratio confirmed the relationship.
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