Prove that points (4,−1,3) & (5,−1,4) lies on same side of the plane x+y+z=7.


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Question Number 45555 by rahul 19 last updated on 14/Oct/18

$P r o v e t h a t p o i n t s (4, - 1, 3) & (5, - 1, 4)$ $l i e s o n s a m e s i d e o f t h e p l a n e x + y + z = 7 .$
Commented byrahul 19 last updated on 14/Oct/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18

	$4 - 1 + 3 - 7 = - 1$ $5 - 1 + 4 - 7 = + 1$ $o p p o s i t e s i d e$ $m i d p o i n t o f (4, - 1, 3) a n d (5, - 1, 4)$ $\frac{4 + 5}{2}, \frac{- 1 - 1}{2}, \frac{3 + 7}{2}$ $(\frac{9}{2}, - 1, \frac{7}{2})$ $n o w p u t (\frac{9}{2}, - 1, \frac{7}{2}) i n x + y + z - 7 = 0$ $\frac{9}{2} - 1 + \frac{7}{2} - 7$ $= 8 - 8$ $= 0 t h a t c o n f i r m p o i n t s a r e o p p o s i t e t o p l s n e s$ $a n d a r e m i r r i r i m a g e . . .$
	Commented bytanmay.chaudhury50@gmail.com last updated on 14/Oct/18

	$y e s q u e s t i o n i s w r o n g . . .$
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