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Distance Formula

The points ...

Question

The points $(- 2, - 1), (1, 0), (4, 3),$ and $(1, 2)$ are the vertices

A

Parallelogram

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B

Rectangle

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C

Square

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D

Both

(1)

&

(2)

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Solution

The correct option is C Parallelogram
$T o i d e n t i f y - w h e t h e r A B C D i s a (i) p a r a l l e l o g r a m, (i i) r e c t a n g l e, (i i i) s q u a r e, (i v) s c a l e n e o n e . S o l u t i o n - (i) W e k n o w t h a t i n a p a r a l l e l o g r m t h e d i a g o n a l s b i s e c t e a c h o t h e r i . e t h e m i d p o i n t s o f A C & B D a r e s a m e . L e t t h e m i d p o i n t o f A C b e P (x, y) . A p p l y i n g m i d p o i n t t h e o r e m, P (x, y) = (\frac{x_{1} + x_{3}}{2}, \frac{y_{1} + y_{3}}{2}) = (\frac{- 2 + 4}{2}, \frac{- 1 + 3}{2}) = (1, 1) . A g a i n l e t t h e m i d p o i n t o f B D b e Q (p, q) . A p p l y i n g m i d p o i n t t h e o r e m, Q (p, q) = (\frac{x_{2} + x_{4}}{2}, \frac{y_{2} + y_{4}}{2}) = (\frac{1 + 1}{2}, \frac{0 + 1}{2}) = (1, 1) . ∴ P (x, y) = Q (p, q) ⟹ t h e m i d p o i n t s o f A C & B D a r e s a m e . S o A B C D i s a p a r a l l e l o g r a m . (i i) I f t h e d i a g o n a l s o f t h e p a r a l l e l o g r a m A B C D a r e e q u a l t h e n i t i s a r e c t a n g l e . i . e A C = B D . A p p l y i n g d i s t a n c e f o r m u l a, A C = \sqrt{{(x_{1} - x_{3})}^{2} + {(y_{1} - y_{3})}^{2}} = \sqrt{{(- 2 - 4)}^{2} + {(- 1 - 3)}^{2}} u n i t s = 2 \sqrt{13} u n i t s . B D = \sqrt{{(x_{2} - x_{4})}^{2} + {(y_{2} - y_{4})}^{2}} = \sqrt{{(1 - 1)}^{2} + {(0 - 2)}^{2}} u n i t s = 2 u n i t s . ∴ A C \neq B D . S o A B C D i s n o t a r e c t a g l e . C o n s e q u e n t l y, i t i s n o t a s q u a r e . A n s - O p t i o n A .$

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