The equation of the sides AB,BC,CA of a triangle ABC are 2x+y=0, x+py=q , x-y=3 respectively.
The point P is (2,3)
Find
1) If P is the centroid, then find p+q
2) If P is the orthocentre, the find p+q
3) If P is the circumcentre, then find p+q.

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \mathrm{AB}:2\mathrm{x}+\mathrm{y}=0 ;\left(\mathrm{i}\right) \\ \mathrm{BC}:\mathrm{x}+\mathrm{py}=\mathrm{q} ;\left(\mathrm{ii}\right) \\ \mathrm{AC}:\mathrm{x}-\mathrm{y}=3 :\left(\mathrm{iii}\right) \\ \left(\mathrm{i}\right)+\left(\mathrm{iii}\right), \mathrm{we} \mathrm{get}, \\ 2\mathrm{x}+\mathrm{y}+\mathrm{x}-\mathrm{y}=0+3 \\ \Rightarrow 3\mathrm{x}=3 \\ \Rightarrow \mathrm{x}=1 \\ \mathrm{Put} \mathrm{x}=1 \mathrm{in} \left(\mathrm{iii}\right), \mathrm{we} \mathrm{get}, \\ 1-\mathrm{y}=3 \\ \Rightarrow \mathrm{y}=-2 \\ \Rightarrow \mathrm{A}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\left(1,-2\right) \\ \left(\mathrm{ii}\right)-\left(\mathrm{iii}\right), \mathrm{we} \mathrm{get}, \\ \mathrm{x}+\mathrm{py}-\mathrm{x}+\mathrm{y}=\mathrm{q}-3 \\ \Rightarrow \mathrm{y}\left(\mathrm{p}+1\right)=\mathrm{q}-3 \\ \Rightarrow \mathrm{y}=\frac{\mathrm{q}-3}{\mathrm{p}+1} \\ \left(\mathrm{ii}\right)+\mathrm{p}\times \left(\mathrm{iii}\right) \\ \mathrm{x}+\mathrm{py}+\mathrm{px}-\mathrm{py}=\mathrm{q}+3\mathrm{p} \\ \Rightarrow \mathrm{x}\left(\mathrm{p}+1\right)=\mathrm{q}+3\mathrm{p} \\ \Rightarrow \mathrm{x}=\frac{3\mathrm{p}+\mathrm{q}}{\mathrm{p}+1} \\ \Rightarrow \mathrm{C}\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)=\left(\frac{3\mathrm{p}+\mathrm{q}}{\mathrm{p}+1},\frac{\mathrm{q}-3}{\mathrm{p}+1}\right) \\ 2\left(\mathrm{ii}\right)-\left(\mathrm{i}\right) \\ 2\mathrm{x}+2\mathrm{py}-2\mathrm{x}-\mathrm{y}=2\mathrm{q} \\ \Rightarrow \mathrm{y}\left(2\mathrm{p}-1\right)=2\mathrm{q} \\ \Rightarrow \mathrm{y}=\frac{2\mathrm{q}}{2\mathrm{p}-1} \\ \mathrm{p}\times \left(\mathrm{i}\right)-\left(\mathrm{ii}\right) \\ 2\mathrm{px}+\mathrm{py}-\mathrm{x}-\mathrm{py}=0-\mathrm{q} \\ \Rightarrow \mathrm{x}\left(2\mathrm{p}-1\right)=-\mathrm{q} \\ \Rightarrow \mathrm{x}=-\frac{\mathrm{q}}{2\mathrm{p}-1} \\ \Rightarrow \mathrm{B}\left({\mathrm{x}}_3,{\mathrm{y}}_3\right)=\left(-\frac{\mathrm{q}}{2\mathrm{p}-1},\frac{2\mathrm{q}}{2\mathrm{p}-1}\right) \\ \mathrm{Case}\left(\mathrm{i}\right) \mathrm{When} \mathrm{P}\left(2,3\right) \mathrm{is} \mathrm{centroid}. \\ \mathrm{A}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\left(1,-2\right) \mathrm{C}\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)=\left(\frac{3\mathrm{p}+\mathrm{q}}{\mathrm{p}+1},\frac{\mathrm{q}-3}{\mathrm{p}+1}\right) \mathrm{B}\left({\mathrm{x}}_3,{\mathrm{y}}_3\right)=\left(-\frac{\mathrm{q}}{2\mathrm{p}-1},\frac{2\mathrm{q}}{2\mathrm{p}-1}\right) \\ \Rightarrow \left(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3}{3},\frac{{\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3}{3}\right)=\left(2,3\right) \\ \Rightarrow \frac{{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3}{3}=3 \\ \Rightarrow {\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3=6 \\ \Rightarrow 1+\frac{3\mathrm{p}+\mathrm{q}}{\mathrm{p}+1}-\frac{\mathrm{q}}{2\mathrm{p}-1}=6 \\ \Rightarrow \frac{3\mathrm{p}+\mathrm{q}}{\mathrm{p}+1}-\frac{\mathrm{q}}{2\mathrm{p}-1}=5 \\ \Rightarrow \left(3\mathrm{p}+\mathrm{q}\right)\left(2\mathrm{p}-1\right)-\mathrm{q}\left(\mathrm{p}+1\right)=5\left(\mathrm{p}+1\right)\left(2\mathrm{p}-1\right) \\ \Rightarrow 6{\mathrm{p}}^2-3\mathrm{p}+2\mathrm{pq}-\mathrm{q}-\mathrm{pq}-\mathrm{q}=10{\mathrm{p}}^2-5\mathrm{p}+10\mathrm{p}-5 \\ \Rightarrow 4{\mathrm{p}}^2-3\mathrm{p}-2\mathrm{q}+\mathrm{pq}=10{\mathrm{p}}^2+5\mathrm{p}-5 \\ \Rightarrow 6{\mathrm{p}}^2+8\mathrm{p}+2\mathrm{q}-\mathrm{pq}=5 ;\left(\mathrm{iv}\right) \\ \mathrm{and} \\ \frac{{\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3}{3}=3 \\ \Rightarrow {\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3=9 \\ \Rightarrow -2+\frac{\mathrm{q}-3}{\mathrm{p}+1}+\frac{2\mathrm{q}}{2\mathrm{p}-1}=9 \\ \Rightarrow \frac{\mathrm{q}-3}{\mathrm{p}+1}+\frac{2\mathrm{q}}{2\mathrm{p}-1}=11 \\ \Rightarrow \left(\mathrm{q}-3\right)\left(2\mathrm{p}-1\right)+2\mathrm{q}\left(\mathrm{p}+1\right)=11\left(\mathrm{p}+1\right)\left(2\mathrm{p}-1\right) \\ \Rightarrow 3\mathrm{pq}-\mathrm{q}-6\mathrm{p}+3+2\mathrm{pq}+2\mathrm{q}=22{\mathrm{p}}^2-11\mathrm{p}+22\mathrm{p}-11 \\ \Rightarrow 5\mathrm{pq}-6\mathrm{p}+\mathrm{q}+3=22{\mathrm{p}}^2+11\mathrm{p}-11 \\ \Rightarrow 22{\mathrm{p}}^2+17\mathrm{p}-\mathrm{q}-5\mathrm{pq}=11 ;\left(\mathrm{v}\right) \\ \mathrm{As} \left(\mathrm{iv}\right) \mathrm{and} \left(\mathrm{v}\right) \mathrm{contains} \mathrm{terms} \mathrm{like} {\mathrm{p}}^2, \mathrm{pq}, \mathrm{p} \mathrm{and} \mathrm{q} \mathrm{hence} \mathrm{it} \mathrm{is} \mathrm{irreducible} \mathrm{to} \\ \mathrm{p}+\mathrm{q}. \mathrm{So} \mathrm{we} \mathrm{cannot} \mathrm{proceed} \mathrm{further}. \end{gathered}$$

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