1. What is the distance of A from B? D from C?
Distance of A from B:
A (-1, 0)
B (2, 0)
[tex]\[\begin{array}{l}d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \\\\d = \sqrt {(2 - {{( - 1)}^2} + {{(0 - 0)}^2}} \\\\d = \sqrt {{{(3)}^2} + {{(0)}^2}} \\\\d = \sqrt {9 + 0} \\\\d = \sqrt 9 \\\\d = 3\end{array}\][/tex]
Distance of D from C:
C (2, 3)
D (-1, 3)
[tex]\[\begin{array}{l}d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \\\\d = \sqrt {{{( - 1 - 2)}^2} + {{(3 - 3)}^2}} \\\\d = \sqrt {{{( - 3)}^2} + {{(0)}^2}} \\\\d = \sqrt {9 + 0} \\\\d = \sqrt 9 \\\\d = 3\end{array}\][/tex]
2.What is the distance of A from D? B from C?
Distance of A from D
A (-1, 0)
D (-1, 3)
[tex]\[\begin{array}{l}d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \\\\d = \sqrt {( - 1 - {{( - 1)}^2} + {{(3 - 0)}^2}} \\\\d = \sqrt {{{(0)}^2} + {{(3)}^2}} \\\\d = \sqrt {0 + 9} \\\\d = \sqrt 9 \\\\d = 3\end{array}\][/tex]
Distance of B from C
B (2, 0)
C (2, 3)
[tex]\[\begin{array}{l}d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \\\\d = \sqrt {{{(2 - 2)}^2} + {{(3 - 0)}^2}} \\\\d = \sqrt {{{(0)}^2} + {{(3)}^2}} \\\\d = \sqrt {0 + 9} \\\\d = \sqrt 9 \\\\d = 3\end{array}\][/tex]
3. Compare the distances
The distances are all equal to 3 units. Each points are 3 units apart
4. What is the figure formed?
Connecting the points will form a square. A square has all sides equal.
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