2 opposite vertices of a square are (5,4) and (1,−6). Find coordinates of remaining two vertices ?


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Question Number 50625 by rahul 19 last updated on 18/Dec/18

$2 o p p o s i t e v e r t i c e s o f a s q u a r e a r e$ $(5, 4) a n d (1, - 6) . F i n d c o o r d i n a t e s$ $o f r e m a i n i n g t w o v e r t i c e s ?$
Commented by $@ty@m last updated on 18/Dec/18

$P l o t t h e s e p o i n t s i n g r a p h p a p e r .$
Commented by rahul 19 last updated on 18/Dec/18

$S i r, c a n y o u p l s h o w y o u r m e t h o d . .$
	Answered by ajfour last updated on 18/Dec/18

	$c e n t e r \equiv (3, - 1)$ $\frac{1}{2} \times d i a g o n a l = \sqrt{29}$ $s l o p e m = \frac{5}{2}$ $s l o p e o f C D = - \frac{2}{5}$ $x_{C} = 3 + \sqrt{29} (\frac{- 5}{\sqrt{29}}) = - 2$ $y_{C} = - 1 + \sqrt{29} (\frac{2}{\sqrt{29}}) = 1$ $x_{D} = 3 - \sqrt{29} (\frac{- 5}{29}) = 8$ $y_{D} = - 1 - \sqrt{29} (\frac{2}{\sqrt{29}}) = - 3$ $C \equiv (- 2, 1)$ $D \equiv (8, - 3) .$
	Commented by rahul 19 last updated on 18/Dec/18
	thanks sir!��
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