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Congruency

Let O be th...

Question

Let $O$ be the origin and let $P Q R$ be an arbitrary triangle. The point $S$ is such that
$O P . O Q + O R . O S = O R . O P + O Q . O S = O Q . O R + O P . O S$
Then the triangle $P Q R$ has $S$ as its

centroid

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orthocentre

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incentre

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circumcentre

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Open in App

Solution

The correct option is C orthocentre

$- - \to O P . - - \to O Q + - - \to O R . - - \to O S = - - \to O R . - - \to O P + - - \to O Q . - - \to O S$
$\Rightarrow (- - \to O Q - - - \to O R) . - - \to O P = (- - \to O Q - - - \to O R) . - - \to O S$
$\Rightarrow (- - \to O Q - - - \to O R) . (- - \to O P - - - \to O S) = 0$
$\Rightarrow - - \to R Q . - \to S P = 0$
$\Rightarrow R Q ⊥ S P ⟶ (A)$
Again,
$- - \to O R . - - \to O P + - - \to O Q . - - \to O S = - - \to O Q . - - \to O R + - - \to O P . - - \to O S$
$\Rightarrow (- - \to O R - - - \to O S) . - - \to O P = (- - \to O R - - - \to O S) - - \to O Q$
$\Rightarrow (- - \to O R - - - \to O S) (- - \to O P - - - \to O Q) = 0$
$\Rightarrow - - \to S R ⊥ - - \to Q P ⟶ (B)$
$∴$ $S$ should be orthocentre [B].

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