Answer
(a) $\sqrt (133)$
(b) $(2\sqrt 2,\frac{3\sqrt 5}{2})$
Work Step by Step
Let $P=(x_{1},y_{1})=(3\sqrt 2,4\sqrt 5)$ and $Q=(x_{2},y_{2})=(\sqrt 2,-\sqrt 5)$
Part (a):
Finding the distance between P and Q,
$d(P,Q)=\sqrt ((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2})$
$d(P,Q)=\sqrt ((\sqrt 2-3\sqrt 2)^{2}+(-\sqrt 5-4\sqrt 5)^{2})$
$d(P,Q)=\sqrt ((-2\sqrt 2)^{2}+(-5\sqrt 5)^{2})$
$d(P,Q)=\sqrt ((4\times2)+(25\times5))$
$d(P,Q)=\sqrt ((8)+(125))$
$d(P,Q)=\sqrt (133)$
Part (b):
The midpoint formula is $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$
Substituting the values, the formula becomes:
$(\frac{3\sqrt 2+\sqrt 2}{2},\frac{4\sqrt 5-\sqrt 5}{2})=(\frac{4\sqrt 2}{2},\frac{3\sqrt 5}{2})=(2\sqrt 2,\frac{3\sqrt 5}{2})$
Therefore, the coordinates of the midpoint are: $(2\sqrt 2,\frac{3\sqrt 5}{2})$
