Question

If the orthocentre, circumcentre of a triangle are (-3,5,2) , (6,2,5) respectively then the centroid of the triangle is
A). $(3,3,4)$
B). $\left( {\dfrac{3}{2},\dfrac{7}{2},\dfrac{9}{2}} \right)$
C). $(9,9,12)$
D). $\left( {\dfrac{9}{2},\dfrac{{ - 3}}{2},\dfrac{3}{2}} \right)$

Answer 1

If H,O and G be the orthocentre, circumcentre and centroid of any triangle. Then, these points are collinear. Further, G divides the line segment HO from H in the ratio $2:1$ internally i.e., $\dfrac{{HG}}{{GO}} = 2:1$

Complete step-by-step answer:
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So we are given the coordinates of H and O all we need to do is to find the coordinates of G
Also here $\dfrac{m}{n} = \dfrac{{HG}}{{GO}} = 2:1$ Now we know that
$x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}}\& y = \dfrac{{m{y_2} + n{y_1}}}{{m + n}}\& z = \dfrac{{m{z_2} + n{z_1}}}{{m + n}}$
Now here

H({x_1},{y_1},{z_1}) = ( - 3,5,2)\\\ O({x_2},{y_2},{z_2}) = (6,2,5)\\\ G(x,y,z) \end{array}$$ So for the value of x We need to put the formula $$x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}}$$ $$\begin{array}{l} \therefore x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}}\\\ \Rightarrow x = \dfrac{{2 \times 6 + 1 \times ( - 3)}}{{2 + 1}}\\\ \Rightarrow x = \dfrac{{12 - 3}}{3}\\\ \Rightarrow x = 3 \end{array}$$ $$\begin{array}{l} \therefore y = \dfrac{{m{y_2} + n{y_1}}}{{m + n}}\\\ \Rightarrow y = \dfrac{{2 \times 2 + 1 \times 5}}{{2 + 1}}\\\ \Rightarrow y = \dfrac{{4 + 5}}{3}\\\ \Rightarrow y = 3 \end{array}$$ And for the last the value of z we will use $$z = \dfrac{{m{z_2} + n{z_1}}}{{m + n}}$$ $$\begin{array}{l} \therefore z = \dfrac{{m{z_2} + n{z_1}}}{{m + n}}\\\ \Rightarrow z = \dfrac{{2 \times 5 + 1 \times 2}}{{2 + 1}}\\\ \Rightarrow z = \dfrac{{12}}{3}\\\ \Rightarrow z = 4 \end{array}$$ **Which means that the centroid is (3,3,4) and hence option A is the correct option here.** **Note:** The relation between orthocentre, circumcentre and centroid is the key point to solve this question, after that we just used the section formula to find the individual coordinates. Be in sound concentration while putting the values of the coordinates as a lot of students make the mistake of putting the wrong coordinate due to some sort of confusion.