Answer
(a) $3\sqrt 55$
(b) $(2\sqrt 7,\frac{7\sqrt 3}{2})$
Work Step by Step
Let $P=(x_{1},y_{1})=(-\sqrt 7,8\sqrt 3)$ and $Q=(x_{2},y_{2})=(5\sqrt 7,-\sqrt 3)$
Part (a):
Finding the distance between P and Q,
$d(P,Q)=\sqrt ((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2})$
$d(P,Q)=\sqrt ((5\sqrt 7-(-\sqrt 7))^{2}+(-\sqrt 3-8\sqrt 3)^{2})$
$d(P,Q)=\sqrt ((6\sqrt 7)^{2}+(-9\sqrt 3)^{2})$
$d(P,Q)=\sqrt ((36\times7)+(81\times3))$
$d(P,Q)=\sqrt ((252)+(243))$
$d(P,Q)=\sqrt (495)$
$d(P,Q)=\sqrt (9\times55)$
$d(P,Q)=\sqrt 9\sqrt 55$
$d(P,Q)=3\sqrt 55$
Part (b):
The midpoint formula is $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$
Substituting the values, the formula becomes:
$(\frac{-\sqrt 7+5\sqrt 7}{2},\frac{8\sqrt 3-\sqrt 3}{2})=(\frac{4\sqrt 7}{2},\frac{7\sqrt 3}{2})=(2\sqrt 7,\frac{7\sqrt 3}{2})$
Therefore, the coordinates of the midpoint are: $(2\sqrt 7,\frac{7\sqrt 3}{2})$
