A) \[\operatorname{y}-2ax=0\]
B) \[y-\left( {{a}^{2}}+1 \right)x=0\]
C) \[~y+x=0\]
D) \[{{\left( \operatorname{a}-1 \right)}^{2}}x-{{\left( \operatorname{a}+1 \right)}^{2}}y=0\]
Correct Answer: D
Solution :
circumcentre \[=(0,0)\] |
Centroid \[=\left( \frac{{{\left( a+1 \right)}^{2}}}{2},\frac{{{\left( a-1 \right)}^{2}}}{2} \right)\] |
We know the circumcentre (O), |
Centroid (G) and orthocentre (H) of a triangle lie on the line joining the O and G. |
Also, \[\frac{HG}{GO}=\frac{2}{1}\]\[\Rightarrow \] Coordinate of orthocentre \[=\frac{3{{(a+1)}^{2}}}{2},\frac{3{{(a-1)}^{2}}}{2}\] |
Now, these coordinates satisfies eqn given in option [d] |
Hence, required eqn of line is\[{{\left( a-1 \right)}^{2}}x-{{\left( a+1 \right)}^{2}}y=0\] |
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