Question

!20 PTS!

Answer 1

For both problems, we can use the section formula.

1) $Z=\left(\frac{(2)(7)+(1)(1)}{3}, \frac{((2)(5)+(1)(2)}{3} \right)=\boxed{(5, 4)}$

2) $P=\left(\frac{(1)(1)+(5)(-9)}{6}, \frac{(1)(8)+(5)(3)}{6} \right)=\boxed{\left(-\frac{22}{3}, \frac{23}{6} \right)}$