Given the points $A=[2;-3]$, $B=[8;1]$, and $T=[6;4]$, find the coordinates of vertex $C$ of triangle $ABC$ with centroid $T$.
Paul's solution:
(1) The median $t_c$ of triangle $ABC$ is the line segment connecting vertex $C$ and the midpoint $S$ of the opposite side (see the figure). Therefore, $S=\frac12(A+B)=[5;-1]$.
(2) $\overrightarrow{ST}=T-S=(1;5)$.
(3) The centroid $T$ divides the median $t_C=SC$ in the ratio $1:2$. Therefore, $\overrightarrow{SC}=2\overrightarrow{ST}=(2;10)$.
(4) $\overrightarrow{SC}=C-S$; therefore $\mathbf{C=}S+\overrightarrow{SC}=\mathbf{[7;9]}.$
Paul's solution is not correct. Where did Paul make a mistake in his procedure?
The mistake is in step (1). The correct calculation of the coordinates of point $S$ is $S=\frac12(B-A)=[3;2]$.
The mistake is in step (2). The vector $\overrightarrow{ST}$ should be $\overrightarrow{ST}=(1;3)$.
The mistake is in step (4). The vector $\overrightarrow{SC}$ is defined as $\overrightarrow{SC}=3\overrightarrow{ST}=(3;15)$.
The mistake is in step (5). The correct calculation of the coordinates of point $C$ is $\mathbf{C=}S-\overrightarrow{SC}=\mathbf{[3;-11]}$.
(1) The median $t_c$ of triangle $ABC$ is the line segment connecting vertex $C$ and the midpoint $S$ of the opposite side (see the figure). Therefore, $S=\frac12(A+B)=[5;-1]$.
(2) $\overrightarrow{ST}=T-S=(1;5)$.
(3) The centroid $T$ divides the median $t_C=SC$ in the ratio $1:2$. Therefore, $\overrightarrow{SC}=3\overrightarrow{ST}=(3;15)$.
(4) $\overrightarrow{SC}=C-S$; therefore $\mathbf{C}=S+\overrightarrow{SC}=\mathbf{[8;14]}.$