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Coordinate plane

This topic has 1 expert reply and 5 member replies
elenaelena Junior | Next Rank: 30 Posts Default Avatar
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Coordinate plane

Post Thu Aug 30, 2012 1:42 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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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    vk_vinayak Legendary Member
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    Post 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.
    ttp://postimage.org/image/y35hzaycb/" target="_blank">

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    elenaelena Junior | Next Rank: 30 Posts Default Avatar
    Joined
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    Posted:
    18 messages
    Post 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.
    ttp://postimage.org/image/y35hzaycb/" target="_blank">
    Thanks a lot for your reply. I will memorize this formula for my future use.

    neelgandham Community Manager
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    Post 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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    \'manpreet singh Master | Next Rank: 500 Posts Default Avatar
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    Post Thu Aug 30, 2012 3:12 am
    √(7-3)^2+(10-5)^2=√16+25=√41

    ans (E)

    elenaelena Junior | Next Rank: 30 Posts Default Avatar
    Joined
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    Posted:
    18 messages
    Post 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!

    Post 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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