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Question

If the centroid and circumcentre of a triangle are $$(3, 3)$$ and $$(6, 2)$$, respectively, then the orthocentre is


A
(3,5)
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B
(3,1)
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C
(3,1)
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D
(9,5)
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Solution

The correct option is A $$(-3, 5)$$
In any triangle, orthocentre, centroid and circumcentre are collinear and centroid divides the line joining orthocentre and circumcenter in the ratio $$2 : 1.$$
Let the orthocentre be $$ (x,y) $$.
Using the section formula, $$\left( \cfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n } ,\cfrac { m{ y }_{ 2}  + n{ y }_{ 1 } }{ m + n }  \right) $$

Substituting $$({ x }_{ 1 },{ y }_{ 1 }) = (x,y) $$ and $$({x }_{ 2 },{ y}_{ 2 }) = (6,2) $$  and $$ m = 2, n = 1 $$ in the section formula, we get the centroid $$ = \left( \cfrac { 2(6)  + 1(x) }{ 2 +1 } ,\cfrac { 2(2) +1(y) }{ 2 + 1 }  \right) = \left(\cfrac { x + 12 }{ 3 } ,\cfrac { y + 4 }{ 3} \right) $$

Given centroid $$ = (3,3) $$

$$ => \left(\cfrac { x + 12 }{ 3 } ,\cfrac { y + 4 }{ 3} \right) = (3,3) $$
$$ => \cfrac { x + 12 }{ 3 } = 3 ; \cfrac { y + 4 }{ 3} = 3 $$
$$ => x + 12 = 9 ; y + 4 = 9 $$
$$=> x = -3 ; y = 5 $$
Hence, orthocentre $$ = (-3,5) $$

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