Question

ivan used coordinate geometry to prove that quadrilateral efgh is a square.

statement\treason

1. quadrilateral efgh is at e (-2, 3), f (1, 6), g (4, 3), and h (1, 0)\t1. given

\t2. \\(overline{ef}\\)\
e (-2, 3) f (1, 6)\
\\(d = \sqrt{(1 + 2)^2 + (6 - 3)^2}\\)\t\\(overline{gh}\\)\
g (4, 3) h (1, 0)\
\\(d = \sqrt{(1 - 4)^2 + (0 - 3)^2}\\)

Explanation:

Answer:

$\overline{EF}$ length: $\sqrt{(1+2)^2+(6-3)^2} = 3\sqrt{2}$
$\overline{GH}$ length: $\sqrt{(1-4)^2+(0-3)^2} = 3\sqrt{2}$
(Additional complete steps would confirm all sides equal and slopes show perpendicular adjacent sides, proving it is a square)