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# 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

**Solution:**

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

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

On using the 3d distance formula, we obtain

PA² = ( x - 3)² + ( y - 4)² + ( z - 5)²

= x² + 9 - 6x + y² + 16 - 8y + z² + 25 - 10z

= x² - 6x + y² - 8 y + z² - 10z + 50

PB² = (x + 1)² + (y - 3)² + (z + 7)²

= x² + 1 + 2x + y² + 9 - 6y + z² + 49 + 14z

= x² + 2x + y² - 6y + z² + 14z + 59

Now, if

PA² + PB² = k², then,

⇒ (x² - 6x + y² - 8 y + z² - 10z + 50) + (x² + 2x + y² - 6y + z² + 14z + 59) = k²

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

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

⇒ (x² + y² + z² - 2x - 7y + 2z) = (k² - 109)/2

Thus, the required equation is (x² + y² + z² - 2x - 7y + 2z) = (k² - 109)/2

NCERT Solutions Class 11 Maths Chapter 12 Exercise ME Question 6

## 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

**Summary:**

The equation of the set of points P such that PA² + PB² = k², where k is a constant and A and B are the points (3, 4, 5) and (- 1, 3, - 7) respectively is (x² + y² + z² - 2x - 7y + 2z) = (k² - 109)/2

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