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Question

Match the following

I. The centroid of the triangle formed by (2,3,−1),(5,6,3),(2,−3,1) isa) (2,2,2)
II. The circumcentre of the triangle formed by (1,2,3),(2,3,1),(3,1,2) isb) (3,1,4)
III. The orthocenter of the triangle formed by (2,1,5),(3,2,3),(4,0,4) isc) (1,1,0)
IV. The incentre of the triangle formed by (0,0,0),(3,0,0),(0,4,0) isd) (3,2,1)
e) (0,0,0)

A
Id,IIa,IIIb,IVc
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B
Ia,IIb,IIIc,IVd
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C
Id,IIe,IIIb,IVc
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D
Id,IIa,IIIe,IVc
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Solution

The correct option is A Id,IIa,IIIb,IVc
I) centroid =(2+5+23,3+633,1+3+13)
=(3,2,1)

II) It is an equilateral triangle as its circumcenter will be its centroid. and centroid is
(1+2+33,2+3+13,3+1+23)=(2,2,2)

III) Let, P(x, y, z) be the orthocenter. So, APBC,BPAC and CPAB.
DRs of AP=(x2,y1,z5)
DRs of BC=(1,2,1)
As, APBC, x22y+2+z5=0
x2y+z=5
check all the remaining options.
Only (3, 1, 4) satisfies it.
So, (3, 1, 4) will be orthocenter

IV) Incentre
a=9+16+0=5
b=4,c=3
Incenter =(ax1+bx2+cx3a+b+c,ay1+by2+cy3a+b+c,az1+az2+az3a+b+c)
=(5×0+4×3+3×03+4+5,5.0+4.0+3.43+4+5,5.0+4.0+3.03+4+5)
=(1,1,0)

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