# A triangle has corners at (-3 ,5 ), (7 ,6 ), and (1 ,-4 ). If the triangle is dilated by a factor of 2/3  about point (-3 ,4 ), how far will its centroid move?

Mar 15, 2016

Thus the new corners of the dilated triangle A'B'C" are
$A ' \left(- 3 , \frac{14}{3}\right) , B ' \left(\frac{11}{3} , \frac{16}{3}\right) , C ' \left(- \frac{1}{3} , - \frac{4}{3}\right)$

#### Explanation:

For each vertex we need to find the dilation of that point by 2/3 around (-3,4):
$A \left(- 3 , 5\right) , B \left(7 , 6\right) , C \left(1 , - 4\right)$
Dilate $B ' = {D}_{\left(- 3 , 4\right) , \frac{2}{3}} \cdot B ' \left(7 , 6\right) \implies \left[7 - \left(- 3\right) , 6 - 4\right] \cdot \frac{2}{3} + \left[- 3 , 4\right] = \left(\frac{11}{3} , \frac{16}{3}\right)$
$A ' = {D}_{\left(- 3 , 4\right) , \frac{2}{3}} \cdot A ' \left(- 3 , 5\right) \implies \left[- 3 - \left(- 3\right) , 5 - 4\right] \cdot \frac{2}{3} + \left[- 3 , 4\right] = \left(- 3 , \frac{14}{3}\right)$
$C ' = {D}_{\left(- 3 , 4\right) , \frac{2}{3}} \cdot C ' \left(1 , - 4\right) \implies \left[1 - \left(- 3\right) , - 4 - 4\right] \cdot \frac{2}{3} + \left[- 3 , 4\right] = \left(- \frac{1}{3} , - \frac{4}{3}\right)$

Thus the new corners of the dilated triangle A'B'C"# are
$A ' \left(- 3 , \frac{14}{3}\right) , B ' \left(\frac{11}{3} , \frac{16}{3}\right) , C ' \left(- \frac{1}{3} , - \frac{4}{3}\right)$