Assertion $(A) :$ If $A (\bar{a}), B (\bar{b}), C (\bar{c})$ are the vertices of a triangle, with circumcentre is $(0, 0)$ then its nine point centre will be $\frac{\bar{a} + \bar{b} + \bar{c}}{2}$
Reason $(R)$ centroid of $Δ A B C$ is $(\frac{\bar{a} + \bar{b} + \bar{c}}{3})$

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Both A & R are individually true abut R is not the correct explanation of A

A is false but R is true

A is true but R is false

Both A & R are individually true and R is the current explanation

answer is B.

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Detailed Solution

$(\frac{\bar{a} + \bar{b} + \bar{c}}{3}) ∴ \frac{\bar{a} + \bar{b} + \bar{c}}{3} = \frac{2 \times 0 + 1. \vec{r}}{2 + 1} \Rightarrow \vec{r} = \vec{a} + \vec{b} + \vec{c} A n d n i n e p o i n t c e n t r e i s t h e m i d p o i n t o f o r t h o c e n t r e a n d c i r c u m c e n t r e n i n e p o i n t c e n t r e = \frac{\vec{r} + \vec{0}}{2} = \frac{\bar{a} + \bar{b} + \bar{c}}{2}$