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Problem 783

Show that the locus of points equidistant from two given points is the plane perpendicular to the line segment joining them at their midpoint.

Expert verified

The locus of points equidistant from two given points A and B can be represented by the equation of a plane: \[(x_B - x_A)x + (y_B - y_A)y + (z_B - z_A)z = \frac{1}{2} \left[(x_B^2 - x_A^2) + (y_B^2 - y_A^2) + (z_B^2 - z_A^2)\right]\]
This plane is perpendicular to the line segment joining A and B, as shown by its normal vector being parallel to the vector \(\overrightarrow{AB}\), and it contains the midpoint M of the line segment. Therefore, the locus of points equidistant from A and B is the plane perpendicular to the line segment joining them at their midpoint M.

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Chapter 44

Show that if a line is perpendicular to one of two parallel planes, then it is perpendicular to the other.

Chapter 44

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Chapter 44

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Chapter 44

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Chapter 44

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