Answer
(a) $4\sqrt 2$
(b) $(-1,-1)$
(c) $y=x$
Work Step by Step
(a) Given $P(1,1), Q(-3,-3)$, we can fine the distance $d(P, Q)=\sqrt {(-3-1)^2+(-3-1)^2}=4\sqrt 2$
(b) the coordinates of the midpoint of the segment PQ can be found as $(\frac{1-3}{2},\frac{1-3}{2})$ or $(-1,-1)$
(c) Assume the equation passing the two points as $y=mx+b$, we have $m=\frac{-3-1}{-3-1}=1$ and $1=1(1)+b$ or $b=0$, thus $y=x$
