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Coordinate plane

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Coordinate plane

by elenaelena » Thu Aug 30, 2012 1:42 am
In the coordinate plane, what is the distance between points (3,5) and (7,10)?

A) 3
B) 3√3
C) 2√5
D) √37
E) √41



Hey all, it seems to be not very hard problem,but I have trouble drawing it that's why I couldn't solve it at the first place. Thanks for help.

Correct answer isE
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by vk_vinayak » Thu Aug 30, 2012 2:13 am
elenaelena wrote:In the coordinate plane, what is the distance between points (3,5) and (7,10)?

A) 3
B) 3√3
C) 2√5
D) √37
E) √41

Hey all, it seems to be not very hard problem,but I have trouble drawing it that's why I couldn't solve it at the first place. Thanks for help.

Correct answer isE
No need to draw in this case. If you know the forumla, you can calculatethe distance between two points

Distance between (x1, y1) and (x2, y2) is SQRT( (y2-y1)^2 + (x2-x1)^2))

Therefore distance between (3,5) and (7,10) is SQRT((5^2) + (4^2) ) = SQRT(41). Choose E.
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by elenaelena » Thu Aug 30, 2012 2:37 am
vk_vinayak wrote:
elenaelena wrote:In the coordinate plane, what is the distance between points (3,5) and (7,10)?

A) 3
B) 3√3
C) 2√5
D) √37
E) √41

Hey all, it seems to be not very hard problem,but I have trouble drawing it that's why I couldn't solve it at the first place. Thanks for help.

Correct answer isE
No need to draw in this case. If you know the forumla, you can calculatethe distance between two points

Distance between (x1, y1) and (x2, y2) is SQRT( (y2-y1)^2 + (x2-x1)^2))

Therefore distance between (3,5) and (7,10) is SQRT((5^2) + (4^2) ) = SQRT(41). Choose E.
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Thanks a lot for your reply. I will memorize this formula for my future use.
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by neelgandham » Thu Aug 30, 2012 3:07 am
Graphical Solution :

Let (3,5) be point A and (7,10) be point B. When plotted on a graph, the line AB looks like the one in the attachement below.

From the graph, AC = 4 Units and BC = 5 units. Since triangle ACB is a right angled triangle,
AB^2 = AC^2 + CB^2 (Pythagorean theorem)
AB^2 = 4^2 + 5^2 = 16 + 25 = 41
Distance between the points = AB = √41
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by \'manpreet singh » Thu Aug 30, 2012 3:12 am
√(7-3)^2+(10-5)^2=√16+25=√41

ans (E)
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by elenaelena » Thu Aug 30, 2012 3:51 am
neelgandham wrote:Graphical Solution :

Let (3,5) be point A and (7,10) be point B. When plotted on a graph, the line AB looks like the one in the attachement below.

From the graph, AC = 4 Units and BC = 5 units. Since triangle ACB is a right angled triangle,
AB^2 = AC^2 + CB^2 (Pythagorean theorem)
AB^2 = 4^2 + 5^2 = 16 + 25 = 41
Distance between the points = AB = √41
Thanks for this picture and very easy solution!
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by Jeff@TargetTestPrep » Wed Nov 08, 2017 4:30 pm
elenaelena wrote:In the coordinate plane, what is the distance between points (3,5) and (7,10)?

A) 3
B) 3√3
C) 2√5
D) √37
E) √41
We can use the distance formula:

d = √[(y2 - y1)^2 + (x2 - x1)^2]

d = √[(10 - 5)^2 + (7 - 3)^2]

d = √(5^2 + 4^2)

d = √(25 + 16) = √41

Answer: E

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[email protected]

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