Let's create triangle LMN with the following coordinates:
- 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).
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Vertices of triangle LMN:
- \( L(0, 0) \)
- \( M(4, 0) \)
- \( N(0, 3) \)
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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)
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Coordinates of points:
- \( L(0, 0) \)
- \( M(4, 0) \)
- \( N(0, 3) \)
- \( Q(0, 1) \)
- \( R(2, 1) \)
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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 \( LM \).
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Use the distance formula to confirm that \( \frac{LQ}{QN} = \frac{MR}{RN} \):
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Calculate the lengths:
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Distance \( LQ \): \[ LQ = \sqrt{(0 - 0)^2 + (1 - 0)^2} = \sqrt{0 + 1} = 1 \]
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Distance \( QN \): \[ QN = \sqrt{(0 - 0)^2 + (3 - 1)^2} = \sqrt{0 + 4} = 2 \] Therefore, \( LQ = 1 \) and \( QN = 2 \).
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Now for \( MR \): \[ MR = \sqrt{(2 - 4)^2 + (1 - 0)^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \]
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And for \( RN \): \[ RN = \sqrt{(0 - 2)^2 + (3 - 1)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]
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Calculate the ratios: \[ \frac{LQ}{QN} = \frac{1}{2} \] \[ \frac{MR}{RN} = \frac{\sqrt{5}}{2\sqrt{2}} \]
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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.