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TAGS: Director  Joined: 07 Jan 2018
Posts: 604
Followers: 7

Kudos [?]: 546  , given: 88

In the xy-coordinate system, the distance between points (2\ [#permalink]
1
KUDOS 00:00

Question Stats: 54% (00:50) correct 45% (02:01) wrong based on 11 sessions
In the xy-coordinate system, the distance between points $$(2\sqrt{3}, -\sqrt{2}) and (5\sqrt{3}, 3\sqrt{2}$$) is approximately

A 4.1
B 5.9
C 6.4
D 7.7
E 8.1
[Reveal] Spoiler: OA Director Joined: 20 Apr 2016
Posts: 828
WE: Engineering (Energy and Utilities)
Followers: 11

Kudos [?]: 606  , given: 124

Re: In the xy-coordinate system, the distance between points (2\ [#permalink]
1
KUDOS
amorphous wrote:
In the xy-coordinate system, the distance between points $$(2\sqrt{3}, -\sqrt{2}) and (5\sqrt{3}, 3\sqrt{2}$$) is approximately

A 4.1
B 5.9
C 6.4
D 7.7
E 8.1

Here we can use the formula:- $$\sqrt{(y2 -y1)^2 + (x2 -x1)^2}$$

Here y2= $$5\sqrt{3}$$; y1 = $$2\sqrt{3}$$

x2 = $$3\sqrt{2}$$ and x1 = $$\sqrt{2}$$

So putting the values in the formula we have:

= $$\sqrt{(5\sqrt{3}-2\sqrt{3})^2+(3\sqrt{2}+\sqrt{2})^2}$$=$$\sqrt{(3\sqrt{3})^2+(4\sqrt{2})^2}$$ = $$\sqrt{59}$$

Now we know $$7^2$$=49 and $$8^2$$=64, so 59 should in between $$7<\sqrt{59}<8$$.

Therefore only D. gives a possible value
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