In an acute triangle `A B C` , if the coordinates of orthocentre `H` are `(4,b)` , of centroid `G` are `(b ,2b-8)` , and of circumcenter `S` are `(-4,

Question

In an acute triangle `A B C` , if the coordinates of orthocentre `H` are `(4,b)` , of centroid `G` are `(b ,2b-8)` , and of circumcenter `S` are `(-4,

In an acute triangle `A B C` , if the coordinates of orthocentre `H` are `(4,b)` , of centroid `G` are `(b ,2b-8)` , and of circumcenter `S` are `(-4,8)` , then `b` cannot be `4` (b) `8` (c) 12 (d) `-12` But no common value of `b` is possible.
A. 4
B. 8
C. 12
D. `-12`

Monjur Alahi

11 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - A::B::C::D
As H (orhtocenter), G (centroid), and C (circumcenter) are collinear we have
`|{:(4,,b,,1),(b,,2b-8,,1),(-4,,8,,1):}|=0`
or `|{:(4,,b,,1),(b-4,,b-8,,0),(-(b+4),,16-2b,,0):}|=0`
or `(b-4)(16-2b)+(b+4)(b-4)=0`
or `2(b-4)(8-b)+(b+4)+(b-8)=0`
or `(8-b)[2b-8)-(b+4)=0`
or `(8-b)(b-12)=0`
Hence `b=8 or 12`, which is wrong because collinearity does not explain centroid, orthocenter, and circumcenter.
Now, H.G, and C are collinear and `HG//GC=2`. Therefore,
`(-8+4)/(3)=b or b=(-4)/(3)`
and `(16+b)/(3)=2b-8 or b=8`
But no common value of b is possible.