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Find the equation of the points P such that PA2 + PB2 = 2K2, where A and B are the points (3, 4, 5) and (−1, 3, −7) respectively. |
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Answer» Let P = (x,y,z),B = (3,4,5) and B = (-1,3-7). Now, PB2 = (x +1)2 + (y – 3)2 + (z + 7)2 and PA2 = (x – 3)2 + (y – 4)2 + (z – 5)2 Given that PA2 + PB2 = 2k2, then (x - 3)2 + (y - 4)2 +(z - 5)2 + (x + 1)2 +(y - 3)2 + (z + 7)2 = 2k2 ⇒ x2 – 6x + 9 + y2 – 8y + 16 + z2 – 10z + 22 + x2 + 2x +1 + y2 – 6y + 9 + z2 +14z + 49 = 2k. ⇒ 2x2 + 2y2 + 2z2 – 4x – 14y + 4z + 109 = 2k2 ⇒ 2x2 + 2y2 + 2z2 – 4x + 4y + 4z = 2k2 -109 |
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