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Math Help - I need Geometry help...(Distance and Midpoints)

  1. #1
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    I need Geometry help...(Distance and Midpoints)

    Hello! I'm new to this forum. I'm specifically here to get some answers out of these eaaasy geometry problems. I'm assuming these problems will be easy from you guys, considering some of the mathematics you guys discuss here. See I didn't have enough time to write my notes down last Friday, so I'm a bit stumped as to what to do.

    Anyways....here it goes...

    Find the coordinates of the missing endpoint given that S is the midpoint of RT.

    43. T(-4,3), S(-1,5)

    44. T(2,8), S(-2, 2)

    54. In the figure, GE bisects BC, and GF bisects AB. GE is a horizontal segment and GF is a vertical segment.
    a. Find the coordinates of points F and E.
    b. Name the coordinates of G and explain how you calculated them.
    c. Describe what relationship, if any, exists between DG and GB. Explain.

    Well, that's all, folks! I finished the other problems last night. I'll be waiting around here for answers , so thanks in advance,
    Last edited by aimerzz4; September 3rd 2007 at 03:54 PM.
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    Quote Originally Posted by aimerzz4 View Post
    Hello! I'm new to this forum. I'm specifically here to get some answers out of these eaaasy geometry problems. I'm assuming these problems will be easy from you guys, considering some of the mathematics you guys discuss here. See I didn't have enough time to write my notes down last Friday, so I'm a bit stumped as to what to do.

    Anyways....here it goes...

    Find the coordinates of the midpoint of a segment having the given endpoints.

    43. T(-4,3), S(-1,5)

    44. T(2,8), S(-2, 2)
    There's a formula for finding the midpoint of a line segment.

    The midpoint between two points, (x_1, y_1) and (x_2,y_2) is given by:

    \mbox {Midpoint} = \left( \frac {x_1 + x_2}{2}, \frac {y_1 + y_2}{2} \right)

    that is, the x-coordinate for the midpoint is the average of the x-coordinates of the two points and the y-coordinate of the midpoint is the average of the y-coordinates of the two points. it makes sense when you think about it. if you average two numbers, you get the number that is midway between them.

    now try using that formula and tell us what you get
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  3. #3
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    The mid-point between two points is simply the average of the coordinates.
    The mid-point between (-4,3) & (6,-9) is (1,-3).
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    Quote Originally Posted by aimerzz4 View Post

    54. In the figure, GE bisects BC, and GF bisects AB. GE is a horizontal segment and GF is a vertical segment.
    a. Find the coordinates of points F and E.
    b. Name the coordinates of G and explain how you calculated them.
    c. Describe what relationship, if any, exists between DG and GB. Explain.
    it would help if you gave the figure for this, if it is at all possible
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    You guys...sorry. I typed the directions off of the wrong line. This is the real directions....

    Find the coordinates of the missing endpoint given that S is the midpoint of RT.

    As for the last problem, I guess I'll scan it right now...so hold on a few minutes....
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    Okay, here's the scan for problem #54:

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    Quote Originally Posted by aimerzz4 View Post
    You guys...sorry. I typed the directions off of the wrong line. This is the real directions....

    Find the coordinates of the missing endpoint given that S is the midpoint of RT.

    As for the last problem, I guess I'll scan it right now...so hold on a few minutes....
    we still use the midpoint formula, but now we equate it to a result.

    For T(-4,3) , S(-1,5)

    Let the coordinates of R be (x_1,y_1)

    Since \mbox {midpoint} \{R,T \} = S we have

    \left( \frac {x_1 - 4}{2}, \frac {y_1 + 3}{2} \right) = (-1,5)

    \Rightarrow \frac {x_1 - 4}{2} = -1 and \frac {y_1 + 3}{2} = 5

    now solve for x_1 and y_1

    the other question in this section is, of course, done the same way
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    Quote Originally Posted by aimerzz4 View Post
    Okay, here's the scan for problem #54:

    a. F is the midpoint of AB and E is the midpoint of BC. i gave you the midpoint formula, try finding them yourself

    b. the x-coordinate of G is the same as that of F (since GF is vertical) and the y-coordinate of G is the same as that of E (since GE is horizontal). i guess you'd have to explain why this is so. so, what do you think? why is this so?

    c. intuition tells me that G is the midpoint of BD (i'm waiting on your answers to know for sure). if so, the relationship is, of course, that \overline {DG} = \overline {GB}
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    I think I found the midpoint for problem 43..... (-5/2, 4)...now I'm confused as to what to do next.
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    Quote Originally Posted by Jhevon View Post
    a. F is the midpoint of AB and E is the midpoint of BC. i gave you the midpoint formula, try finding them yourself

    b. the x-coordinate of G is the same as that of F (since GF is vertical) and the y-coordinate of G is the same as that of E (since GE is horizontal). i guess you'd have to explain why this is so. so, what do you think? why is this so?

    c. intuition tells me that G is the midpoint of BD (i'm waiting on your answers to know for sure). if so, the relationship is, of course, that \overline {DG} = \overline {GB}
    OK....for F I got F(4, 6)....and for G I got G(4, 4)..I got 4, 4 by using the midpoint formula.
    BD have a relationship because they're both the endpoints of G??
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    Quote Originally Posted by aimerzz4 View Post
    OK....for F I got F(4, 6)....and for G I got G(4, 4)..I got 4, 4 by using the midpoint formula.
    BD have a relationship because they're both the endpoints of G??
    seems correct. i already told you what the relationship between the two would be if you found that G was the midpoint. (in math, a "relationship" is usually represented by an equation of some sort)
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    Quote Originally Posted by Jhevon View Post
    we still use the midpoint formula, but now we equate it to a result.

    For T(-4,3) , S(-1,5)

    Let the coordinates of R be (x_1,y_1)

    Since \mbox {midpoint} \{R,T \} = S we have

    \left( \frac {x_1 - 4}{2}, \frac {y_1 + 3}{2} \right) = (-1,5)

    \Rightarrow \frac {x_1 - 4}{2} = -1 and \frac {y_1 + 3}{2} = 5

    now solve for x_1 and y_1

    the other question in this section is, of course, done the same way
    Still confused about this one.....I used the midpoint formula for T and S and I got -5/2, 4.....
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    I feel so stupid, but how do I divide 2 and x??



    I can't remember...I just got back to school and a bit rusty.....
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  14. #14
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    Quote Originally Posted by aimerzz4 View Post
    Still confused about this one.....I used the midpoint formula for T and S and I got -5/2, 4.....
    that is wrong. i gave you the formula to use. all you have to do is solve for x_1 and y_1 in the equations i gave you. R = (x_1,y_1) and they want you to find R

    remember, S is the midpoint of RT. so we need to do the midpoint of R and T not S and T. when we do the midpoint for R and T we get S, which is what i did.
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  15. #15
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    Yeah I just realized that. How do I solve this when x1 is not given....
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