Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Distance between two points - Set of points
Last updated at May 29, 2023 by Teachoo
Misc 6 If A and B be the points (3, 4, 5) and (β1, 3, β7), respectively, find the equation of the set of points P such that PA2 + PB2 = k2, where k is a constant. Given points A (3, 4, 5) & B ( β 1, 3, β7) Let Point P be (x, y, z,) We need to point equation of points P, such that PA2 + PB2 = k2 Calculating PA2 P (x, y, z) , A (3, 4, 5) PA = β((x2βx1)2+(y2βy1)2+(z2 βz1)2) Here, x1 = x, y1 = y, z1 = z x2 = 3, y2 = 4, z2 = 5 PA = β((3βx)2+(4βy)2+(5βz)2) Squaring both sides (PA)2 = β((3βx)2+(4βy)2+(5βz)2) = (3 β x)2 + (4 β y)2 + (5 β z)2 = (3)2 + (x)2 β 2(3) (x) + (4)2 + y2 β 2(4)(y) + (5)2 + (z)2 β 2(5)(z) = 9 + x2 β 6x + 16 + y2 β 8y + 25 + z2 β 10z = x2 + y2 β 6x β 8y β 10z + 9 + 16 + 25 = x2 + y2 + z2 β 6x β 8y β 10z + 50 Calculating PB2 P (x, y, z) , B ( β1, 3 , β7) PB = β((x2βx1)2+(y2βy1)2+(z2 βz1)2) Here, x1 = x, y1 = y, z1 = z x2 = β1, y2 = 3, z2 = β7 PB = β((β1βx)2+(3βy)2+(β7βz)2) PB = β((β1)2 (1+π₯)2+(3βy)2+(β1)2(7βz)2) PB = β((1+π₯)2+(3βy)2+(7βz)2) Squaring both sides (PB)2 = (β((1+π₯)2+(3βy)2+(7+z)2))^2 (PB)2 = (1 + x)2 + (3 β y)2 + (7 + z)2 (PB)2 = (1)2 + x2 + 2(1)(x) + (3)2 + (y)2 β 2(3)(y) + (7)2 + z2 + 2(7)(z) = 1 + x2 + 2x + 9 + y2 β 6y + 49 + z2 + 14z = x2 + y2 + z2 + 2x β 6y + 14z + 1 + 9 + 49 = x2 + y2 + z2 + 2x β 6y + 14z + 59 Putting value of (PA)2 & (PB)2 in (1) (PA)2 + (PB)2 = k2 (x2 + y2 + z2 β 6x β 8y β 10z + 50) + (x2 + y2 + z2 + 2x β 6y + 14z + 59) = k2 2x2 + 2y2 + 2z2 β 6x + 2x β 8y β 6y β 10z + 14z + 50 + 59 = k2 2x2 + 2y2 + 2z2 β 4x β 14y + 4z + 109 = k2 2(x2 + y2 + z2) β 4x β 14y + 4z + 109 β k2 = 0 2(x2 + y2 + z2) β 4x β 14y + 4z = k2 β 109 Dividing by 2 both sides (2(π₯^2 + π¦^2 + π§^2))/2β4π₯/2β14π¦/2+4π§/2=(π^2 β 109)/2 x2 + y2 + z2 β 2x β 7y + 2z = (π^2 β 109)/2 Thus the required equation is x2 + y2 + z2 β 2x β 7y + 2z = (π^π β πππ)/π