To determine if the lines through the two sets of points intersect, we need to analyze their slope and y-intercept.
First Set of Points:
- Points: (-3, 8), (-2, 10)
- The slope (m) can be calculated using the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
- Using the points: \[ m = \frac{10 - 8}{-2 + 3} = \frac{2}{1} = 2 \]
- Now we take one point, say (-3, 8), and plug it into the slope-intercept form \( y = mx + b \) to find the y-intercept (b): \[ 8 = 2(-3) + b \] \[ 8 = -6 + b \] \[ b = 14 \]
- The equation of the first line is \( y = 2x + 14 \).
Second Set of Points:
- Points: (12, 0), (14, 1), (16, 2), etc.
- Using the points (12, 0) and (14, 1): \[ m = \frac{1 - 0}{14 - 12} = \frac{1}{2} \]
- Using the point (12, 0) to find the y-intercept: \[ 0 = \frac{1}{2}(12) + b \] \[ 0 = 6 + b \] \[ b = -6 \]
- The equation of the second line is \( y = \frac{1}{2}x - 6 \).
Determine Intersection:
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Set the equations equal to find the x-coordinate of the intersection: \[ 2x + 14 = \frac{1}{2}x - 6 \]
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Multiply through by 2 to eliminate the fraction: \[ 4x + 28 = x - 12 \] \[ 4x - x = -12 - 28 \] \[ 3x = -40 \] \[ x = -\frac{40}{3} \]
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Substitute \( x = -\frac{40}{3} \) into either equation to find the y-coordinate: For the first line: \[ y = 2(-\frac{40}{3}) + 14 = -\frac{80}{3} + 14 = -\frac{80}{3} + \frac{42}{3} = -\frac{38}{3} \] So the intersection point is \((- \frac{40}{3}, - \frac{38}{3})\).
Since we determined two lines with distinct slopes (2 and \( \frac{1}{2} \)), they are not parallel. Thus, they are intersecting lines.
Therefore, the answer is:
Yes...these lines are not parallel nor perpendicular.
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