Answer
a) $\sqrt{29}$
b) $\left(\frac{1}{2},-4\right)$
c) $-\dfrac{2}{5}$
Work Step by Step
RECALL:
(1) The distance $d$ from $P_1(x_1,y_1)$ to $P_2(x_2,y_2)$ is given by the formula
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.
(2) The midpoint $M$ of the line segment from $P_1(x_1,y_1)$ to $P_2(x_2,y_2)$ is:
$M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$.
(3) The slope $m$ of the line that passes through $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ is given by the formula
$m=\dfrac{y_2-y_1}{x_2-x_1}$.
Use the formulas above to obtain:
a) $d=\sqrt{(3-(-2))^2+(-5-(-3))^2}=\sqrt{25+4}=\sqrt{29}$
b) $M=\left(\frac{-2+3}{2},\frac{-3+(-5)}{2}\right)=\left(\frac{1}{2},-4\right)$.
c) $m=\dfrac{-5-(-3)}{3-(-2)}=\dfrac{-2}{5}$
