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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.
See the step by step solution

## Step 1: Define the given points and their midpoint

Let A and B be two given points in space, with coordinates $$(x_A, y_A, z_A)$$ and $$(x_B, y_B, z_B)$$. We can find their midpoint, M, by taking the average of each coordinate: $M(x_M, y_M, z_M) = \left(\frac{x_A+x_B}{2}, \frac{y_A+y_B}{2}, \frac{z_A+z_B}{2}\right)$

## Step 2: Define the locus of points equidistant from the given points

Let P be a point $$(x, y, z)$$ in space equidistant from points A and B. Therefore, the distance between P and A equals the distance between P and B: $PA = PB$ Using the distance formula, we can write: $\sqrt{(x-x_A)^2 + (y-y_A)^2 + (z-z_A)^2} = \sqrt{(x-x_B)^2 + (y-y_B)^2 + (z-z_B)^2}$

## Step 3: Show that the locus of points is a plane

Squaring both sides of the equation and simplifying, we get: $(x-x_A)^2 - (x-x_B)^2 + (y-y_A)^2 - (y-y_B)^2 + (z-z_A)^2 - (z-z_B)^2 = 0$ Rearranging and simplifying further, we obtain the general 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 equation represents the locus of points equidistant from A and B, which is a plane.

## Step 4: Prove that the plane is perpendicular to the line segment joining the given points

The normal vector, $$\bold{n}$$, to the plane can be obtained from the coefficients of x, y, and z in the general equation of the plane: $\bold{n} = (x_B-x_A, y_B-y_A, z_B-z_A)$ We can see that the normal vector is parallel to the vector $$\overrightarrow{AB}$$, and therefore, the plane is perpendicular to the line segment joining points A and B.

## Step 5: Verify that the midpoint of the line segment lies on the plane

To show that the midpoint, M, of the line segment lies on the plane, we substitute M's coordinates into the equation of the plane: $(x_B - x_A)\frac{x_A+x_B}{2} + (y_B - y_A)\frac{y_A+y_B}{2} + (z_B - z_A)\frac{z_A+z_B}{2} = \frac{1}{2} \left[(x_B^2 - x_A^2) + (y_B^2 - y_A^2) + (z_B^2 - z_A^2)\right]$ Simplifying the left side, we can see that it equals the right side, confirming that M lies on the plane. Therefore, we have shown that the locus of points equidistant from two given points A and B is the plane perpendicular to the line segment joining them at their midpoint M.

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