Question

Find the distance between (–3, –3) and (6, 12). Round the answer to the nearest hundredth, if necessary.

Answer 1

$A=(x_A;y_A)\ \ \ and\ \ \ B=(x_B;y_B)\\\\ \Rightarrow\ \ \ |AB|= \sqrt{(x_A-x_B)^2+(y_A-y_B)^2} \\----------------------\\\\A=(-3;-3);\ \ \ B=(6;12)\\\\|AB|= \sqrt{(-3-6)^2+(-3-12)^2} =\\\\.\ \ \ \ \ \ = \sqrt{81+225} = \sqrt{306} =3 \sqrt{34} \approx17.49$

Answer 2

$(-3, -3), \ \ (6, 12)\\ \\distance \ formula : \\ \\d =\sqrt{(x_{2}-x_{1})^2 +(y_{2}-y_{1})^2} \\ \\d =\sqrt{(6-(-3))^2 +(12-(-3) )^2} =\sqrt{(6+3)^2 +(12+3 )^2} =\sqrt{9^2+15^2}=\\ \\=\sqrt{81+225}=\sqrt{306}=\sqrt{9\cdot 34}=3\sqrt{34}=3\cdot 5.83= 17.49$