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જો ત્રિકોણ $ABC$ માં $ A \equiv (1, 10) $, પરિકેન્દ્ર $\equiv$ $\left( { - \,\,{\textstyle{1 \over 3}}\,\,,\,\,{\textstyle{2 \over 3}}} \right)$ અને લંબકેન્દ્ર $\equiv$ $\left( {{\textstyle{{11} \over 3}}\,\,,\,\,{\textstyle{4 \over 3}}} \right)$ હોય તો બિંદુ $A$ ની સામેની બાજુના મધ્યબિંદુના યામો મેળવો
$(1, - 11/3)$
$(1, 5)$
$(1, - 3)$
$(1, 6)$
Solution
We know that,
The orthocentre,centroid and circumcentre of any triangle are collinear.And the centroid divides
the distance from othocentre to circumcentre in the ratio $2: 1 .$
Let the centroid be $G(x, y)$, its coordinates can be found using the section formula.
$\therefore(x, y) \equiv\left(\frac{\frac{-2}{3}+\frac{11}{3}}{3}, \frac{\frac{4}{3}+\frac{4}{3}}{3}\right) \equiv\left(1, \frac{8}{9}\right)$
Also, the centroid $(G)$ divides the medians $(A D)$ in the ratio 2: 1
Let the coordinates of $D$ be $(h, k)$
$\therefore 1=\frac{2 h+1}{3}$ and $\frac{8}{9}=\frac{2 k+10}{3}$
$h=1$ and $8=6 k+30$
and $k=\frac{-11}{3}$
$\therefore D(h, k)=\left(1, \frac{-11}{3}\right)$
