Question

Here are the endpoints of the segments BC, FG, and JK.
B, −67

Answer 1

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{-6}~,~\stackrel{y_1}{7})\qquad C(\stackrel{x_2}{-4}~,~\stackrel{y_2}{4})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ BC=√([-4 - (-6)]^2 + [4 - 7]^2)\implies BC=√((-4+6)^2+(-3)^2) \\\\\\ BC=√(2^2+(-3)^2)\implies \boxed{BC=√(13)} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points}`

`F(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-4})\qquad G(\stackrel{x_2}{1}~,~\stackrel{y_2}{-2})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ FG=√([1 - (-2)]^2 + [-2 - (-4)]^2)\implies FG=√((1+2)^2+(-2+4)^2) \\\\\\ FG=√(9+4)\implies \boxed{FG=√(13)} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ J(\stackrel{x_1}{4}~,~\stackrel{y_1}{2})\qquad K(\stackrel{x_2}{5}~,~\stackrel{y_2}{-2})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2)`

`JK=√([5 - 4]^2 + [-2 - 2]^2)\implies JK=√(1^2+(-4)^2)\implies \boxed{JK=√(17)} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \overline{BC}\cong \overline{FG}~\hfill`