Learn Miscellaneous Exercise 12 with Free Lessons & Tips

The coordinates of points A and B are given as (3, 4, 5) and (–1, 3, –7) respectively.

Let the coordinates of point P be (x, y, z).

On using distance formula, we obtain

Now, if PA² + PB² = k², then

Thus, the required equation is.

The three vertices of a parallelogram ABCD are given as A (3, –1, 2), B (1, 2, –4), and C (–1, 1, 2). Let the coordinates of the fourth vertex be D (x, y, z).

We know that the diagonals of a parallelogram bisect each other.

Therefore, in parallelogram ABCD, AC and BD bisect each other.

∴Mid-point of AC = Mid-point of BD

⇒ x = 1, y = –2, and z = 8

Thus, the coordinates of the fourth vertex are (1, –2, 8).

Let AD, BE, and CF be the medians of the given triangle ABC.

Since AD is the median, D is the mid-point of BC.

∴Coordinates of point D == (3, 2, 0)

Thus, the lengths of the medians of ΔABC are.

It is known that the coordinates of the centroid of the triangle, whose vertices are (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃), are.

Therefore, coordinates of the centroid of ΔPQR

It is given that origin is the centroid of ΔPQR.

Thus, the respective values of a, b, and c are

If a point is on the y-axis, then x-coordinate and the z-coordinate of the point are zero.

Let A (0, b, 0) be the point on the y-axis at a distance of from point P (3, –2, 5). Accordingly,

Thus, the coordinates of the required points are (0, 2, 0) and (0, –6, 0).