To find the distance between the lines defined by the points given, we need to first determine the equations of each line and then determine if any two lines are parallel or intersecting. If they are parallel, we can calculate the distance between them.
Line Equations
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Line A: Points (-10, 10) and (-5, 7)
Slope (m) of Line A: \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 10}{-5 + 10} = \frac{-3}{5} \]
Using point-slope form \(y - y_1 = m(x - x_1)\): \[ y - 10 = -\frac{3}{5}(x + 10) \] Rearranging gives the equation of Line A: \[ y = -\frac{3}{5}x - 6 \]
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Line B: Points (-2, 5) and (-10, -10)
Slope of Line B: \[ m = \frac{-10 - 5}{-10 + 2} = \frac{-15}{-8} = \frac{15}{8} \]
Using point-slope form: \[ y - 5 = \frac{15}{8}(x + 2) \] Rearranging gives the equation of Line B: \[ y = \frac{15}{8}x + \frac{55}{8} \]
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Line C: Points (1, 2) and (-4, 10)
Slope of Line C: \[ m = \frac{10 - 2}{-4 - 1} = \frac{8}{-5} = -\frac{8}{5} \]
Using point-slope form: \[ y - 2 = -\frac{8}{5}(x - 1) \] Rearranging gives the equation of Line C: \[ y = -\frac{8}{5}x + \frac{18}{5} \]
Checking Line Parallelism
Next, we check if any two lines are parallel by comparing their slopes.
- Slope of Line A: \(-\frac{3}{5}\)
- Slope of Line B: \(\frac{15}{8}\)
- Slope of Line C: \(-\frac{8}{5}\)
None of the slopes are equal; therefore, none of the lines are parallel.
Finding Distance Between Lines
Since the lines are not parallel, we can select any two lines to calculate the distance between a point on one line to the other line.
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Distance from a point on Line A to Line B:
- Point on Line A: \((-10, 10)\)
Using the formula for the distance \(d\) from point \((x_0, y_0)\) to line \(Ax + By + C = 0\): \[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Convert Line B into standard form: \[ y - \frac{15}{8}x - \frac{55}{8} = 0 \implies -\frac{15}{8}x + y - \frac{55}{8} = 0 \] Thus, \(A = -\frac{15}{8}, B = 1, C = -\frac{55}{8}\).
Substituting the coordinates of point \((-10, 10)\): \[ d = \frac{\left| -\frac{15}{8}(-10) + 10 - \frac{55}{8} \right|}{\sqrt{(-\frac{15}{8})^2 + 1^2}} = \frac{\left| \frac{150}{8} + 10 - \frac{55}{8} \right|}{\sqrt{\frac{225}{64} + 1}} \] Converting \(10 = \frac{80}{8}\): \[ = \frac{\left| \frac{150 + 80 - 55}{8} \right|}{\sqrt{\frac{225 + 64}{64}}} = \frac{\left| \frac{175}{8} \right|}{\sqrt{\frac{289}{64}}} = \frac{\frac{175}{8}}{\frac{17}{8}} = \frac{175}{17} \]
Conclusion
Calculating the numerical value gives: \[ d \approx 10.29 \] However, if we take the integer part which is the typical requirement for such problems, the final answer can be approximated as: \[ \text{Distance } \approx 10 \]
Since you requested an integer answer for the distance of Line C, the best approach in this context is to round or take the floor value. Therefore, the integer distance we initially calculated rounded down is:
Distance = 10.