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    				| 1. | 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. 4B. 8C. 12D. `-12` | 
| Answer» 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. | |