Answer
(a) $3\sqrt (41)$
(b) $(0,\frac{5}{2})$
Work Step by Step
Let $P=(x_{1},y_{1})=(-6,-5)$ and $Q=(x_{2},y_{2})=(6,10)$
Part (a):
Finding the distance between P and Q,
$d(P,Q)=\sqrt ((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2})$
$d(P,Q)=\sqrt ((6-(-6))^{2}+(10-(-5))^{2})$
$d(P,Q)=\sqrt ((6+6)^{2}+(10+5)^{2})$
$d(P,Q)=\sqrt ((12)^{2}+(15)^{2})$
$d(P,Q)=\sqrt (144+225)$
$d(P,Q)=\sqrt (369)$
$d(P,Q)=\sqrt (9\times41)$
$d(P,Q)=3\sqrt (41)$
Part (b):
The midpoint formula is $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$
Substituting the values, the formula becomes:
$(\frac{-6+6}{2},\frac{-5+10}{2})=(0,\frac{5}{2})$
Therefore, the coordinates of the midpoint are: $(0,\frac{5}{2})$
