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Q.Find the equation of set of the points P such that its distances from the points A(3,4,−5)A(3, 4, -5) and B(−2,1,4)B(-2, 1, 4) are equal. OR Show that the points (−2,3,5)(-2, 3, 5), (1,2,3)(1, 2, 3) and (7,0,−1)(7, 0, -1) are Collinear.

Himachal HpboseHPBOSE Himachal Pradesh Class 11 Board Exam 2026Subjective· 3mImportance★★★★★
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Equating PA2=PB2PA^2=PB^2 for P(x,y,z)P(x,y,z), A(3,4,−5)A(3,4,-5), B(−2,1,4)B(-2,1,4) and simplifying gives 10x+6y−18z−29=010x+6y-18z-29=0.

Let P(x,y,z)P(x,y,z) be a point equidistant from A(3,4,−5)A(3,4,-5) and B(−2,1,4)B(-2,1,4), i.e. PA=PBPA=PB, so PA2=PB2PA^2=PB^2.

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

PB2=(x+2)2+(y−1)2+(z−4)2PB^2 = (x+2)^2+(y-1)^2+(z-4)^2

Expanding:

PA2=x2+y2+z2−6x−8y+10z+50PA^2 = x^2+y^2+z^2 -6x-8y+10z+50

PB2=x2+y2+z2+4x−2y−8z+21PB^2 = x^2+y^2+z^2 +4x-2y-8z+21

Setting PA2=PB2PA^2=PB^2, the x2+y2+z2x^2+y^2+z^2 terms cancel:

−6x−8y+10z+50=4x−2y−8z+21-6x-8y+10z+50 = 4x-2y-8z+21

−6x−4x−8y+2y+10z+8z+50−21=0-6x-4x-8y+2y+10z+8z+50-21=0

−10x−6y+18z+29=0-10x-6y+18z+29=0

Multiplying through by −1-1:

10x+6y−18z−29=010x+6y-18z-29=0

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