If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that PA2 + PB2 = k2, where k is a constant.

Asked by Pragya Singh | 11 months ago |  145

##### Solution :-

Given:

The points A (3, 4, 5) and B (-1, 3, -7)

x1 = 3, y1 = 4, z1 = 5;

x2 = -1, y2 = 3, z2 = -7;

PA2 + PB2 = k2 ……….(1)

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

Now by using distance formula,

We know that the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by

$$PQ= \sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}$$

PA = $$\sqrt{(3 – x)^2 + (4 – y)^2 + (5 – z)^2}$$

PB = $$\sqrt{(-1 – x)^2 + (3 – y)^2 + (-7 – z)^2}$$

Now, substituting these values in (1), we have

[(3 – x)2 + (4 – y)2 + (5 – z)2] +

[(-1 – x)2 + (3 – y)2 + (-7 – z)2] = k2

[(9 + x2 – 6x) + (16 + y2 – 8y) + (25 + z2 – 10z)] +

[(1 + x2 + 2x) + (9 + y2 – 6y) + (49 + z2 + 14z)] = k2

9 + x2 – 6x + 16 + y2 – 8y + 25 + z2 – 10z + 1 +

x2 + 2x + 9 + y2 – 6y + 49 + z2 + 14z = k2

2x2 + 2y2 + 2z2 – 4x – 14y + 4z + 109 = k2

2x2 + 2y2 + 2z2 – 4x – 14y + 4z = k2 – 109

2 (x2 + y2 + z2 – 2x – 7y + 2z) = k2 – 109

(x2 + y2 + z2 – 2x – 7y + 2z) '

$$\dfrac{ (k^2 – 109)}{2}$$

Hence, the required equation is (x2 + y2 + z2 – 2x – 7y + 2z)

$$\dfrac{ (k^2 – 109)}{2}$$

Answered by Abhisek | 11 months ago

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