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Example 8 Find the equation of the set of the points P such that its distances from the points A (3, 4, –5) and B (– 2, 1, 4) are equal. Given A (3, 4, −5) & B ( –2, 1, 4) Let point P be (x, y, z,) Given PA = PB Calculating PA PA = √((x2−x1)2+(y2−y1)2+(z2 −z1)2) Here, x1 = x, y1 = y, z1 = z x2 = 3, y2 = 4, z2 = −5 PA = √((3−𝑥)2+(4−𝑦)2+(−5−𝑧)2) = √((3−𝑥)2+(4−𝑦)2+(5+𝑧)2) = √((3)2+(𝑥)2−2(3)(𝑥)+(4)2+𝑦2−2(4)(𝑦)+(5)2+(𝑧)2+2(5)(𝑧) ) = √(9+𝑥2−6𝑥+16+𝑦2−8𝑦+25+𝑧2+10𝑧) = √(𝑥2+𝑦2+𝑧2−6𝑥−8𝑦+10𝑧+9+16+25) = √(𝑥2+𝑦2+𝑧2−6𝑥−8𝑦+10𝑧+50) Calculating PB P (x, y, z) B (–2, 1, 4) PB = √((x2−x1)2+(y2−y1)2+(z2 −z1)2) Here, x1 = x, y1 = y, z1 = z x2 = –2, y2 = 1, z2 = 4 PB = √((−2−𝑥)2+(1−𝑦)2+(4−𝑧)2) = √((2+𝑥)2+(1−𝑦)2+(4−𝑧)2) = √((2)2+(𝑥)2+2(2)(𝑥)+(1)2+𝑦2−2(1)(𝑦)+42+𝑧2−2(4)(𝑧) ) = √(4+𝑥2+4𝑥+1+𝑦2−2𝑦+16+𝑧2−8𝑧) = √(𝑥2+𝑦2+𝑧2+4𝑥−2𝑦−8𝑧+21) Now, given that PA = PB √(𝑥2+𝑦2+𝑧2−6𝑥−8𝑦+10𝑧+40) = √(𝑥2+𝑦2+𝑧2+4𝑥−2𝑦+8𝑧+21) Squaring both sides (√(𝑥2+𝑦2+𝑧2−6𝑥−8𝑦+10𝑧+40))2 = (√(𝑥2+𝑦2+𝑧2+4𝑥−2𝑦+8𝑧+21))2 𝑥2+𝑦2+𝑧2−6𝑥−8𝑦+10𝑧+40 = 𝑥2+𝑦2+𝑧2+4𝑥−2𝑦+8𝑧+21 𝑥2+𝑦2+𝑧2−6𝑥−8𝑦+10𝑧+40 – 𝑥2−𝑦2−𝑧2+4𝑥+2𝑦+8𝑧−21=0 𝑥2−𝑥2+𝑦2+𝑦2+𝑧2−𝑧2−6𝑥−4𝑥+8𝑦+2y+10z+8z+40−21=0 0 + 0 + 0 – 10x – 6y + 18z + 29 = 0 –10x – 6y + 18z + 29 = 0 0 = 10x + 6y – 18z – 29 = 0 10x + 6y – 18z – 29 = 0 which is the required equation

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.