## Thomas' Calculus 13th Edition

$\left\{\begin{array}{l} \Delta x=2\\ \\ \Delta y=-4 \end{array}\right., \qquad$distance = $2\sqrt{5}$
$A=(x_{1},y_{1})=(-3,2)$ $B=(x_{2},y_{2})=(-1,-2)$ The increments in the coordinates are calculated as $\left\{\begin{array}{l} \Delta x=x_{2}-x_{1}\\ \Delta y=y_{2}-y_{1} \end{array}\right.$ $\left\{\begin{array}{l} \Delta x=-1-(-3)=2\\ \\ \Delta y=-2-2=-4 \end{array}\right.$ The distance formula: $\quad d=\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}$ $d=\sqrt{2^{2}+(-4)^{2}}=\sqrt{4+16}=\sqrt{20}=\sqrt{4\cdot 5}=2\sqrt{5}$