Question

Two vertices of a ∆ABC are given by A(2, 3) and B(-2,1) and it's centroid is G(1,2/3). Find the coordinates of the third vertex C of the ∆ABC.​

Answer 1

We know that

$\frac {{x}_{1} + {x}_{2} + {x}_{3}}{3} ~ , ~ \frac {{y}_{1} + {y}_{2} + {y}_{3}}{3} \\\\\ A (2,3) \\\\\ B(-2,1) \\\\\ C(x,y) \\\\\ G(1, \frac{2}{3})$

$G(x,y) = \frac {{x}_{1} + {x}_{2} + {x}_{3}}{3} ~ , ~ \frac {{y}_{1} + {y}_{2} + {y}_{3}}{3}$

$(1, \frac {2}{3}) = \frac {2+(-2)+x}{3} ~,~ \frac{3+1+y}{3}$

So ,

$1 = \frac {2-2+x}{3} \\\\\ 1 = \frac {x}{3} \\\\\ x = 3$

&

$\frac{2} {{\cancel{3}}} = \frac {3+1+y} {{\cancel{3}}} \\\\\ 2 = 4 + y \\\\\ -2 = y$

So the coordinate of c is ( 3 , -2 ) .