# Geometry: Measuring Segments

##### Junior Member
Find the value of the variable and GH if H is between G and I.

GI = 5b + 2, HI = 4b – 5, HI = 3

My work:

4b - 5 = 3
5 + 3 = 8
4b = 8
b = 2

GI = 5b + 2, and since b = 2
GI = 5(2) + 2
GI = 12

I don’t know how to get to GH.

#### Subhotosh Khan

##### Super Moderator
Staff member
Find the value of the variable and GH if H is between G and I.

GI = 5b + 2, HI = 4b – 5, HI = 3

My work:

4b - 5 = 3
5 + 3 = 8
4b = 8
b = 2

GI = 5b + 2, and since b = 2
GI = 5(2) + 2
GI = 12

I don’t know how to get to GH.
Is there a drawing showing G, H & I are collinear (in one st. line)?

#### ksdhart2

##### Senior Member
.
Find the value of the variable and GH if H is between G and I.

GI = 5b + 2, HI = 4b – 5, HI = 3
Please double check to make sure you copied the problem down correctly. Taken exactly as you wrote it here, the question makes no sense and is unanswerable. The line segment from the point G to the point I (denoted GI) is given by y = 12. This is a horizontal line. Likewise, the line segment from the point H to the point I (denoted HI) is given by y = 3. This is also a horizontal line. These two line segments must share one end point (the point I), but how can this be when the two lines are both horizontal and thus parallel to one another?

##### Junior Member
Is there a drawing showing G, H & I are collinear (in one st. line)?
No drawing.

##### Junior Member
.

Please double check to make sure you copied the problem down correctly. Taken exactly as you wrote it here, the question makes no sense and is unanswerable. The line segment from the point G to the point I (denoted GI) is given by y = 12. This is a horizontal line. Likewise, the line segment from the point H to the point I (denoted HI) is given by y = 3. This is also a horizontal line. These two line segments must share one end point (the point I), but how can this be when the two lines are both horizontal and thus parallel to one another?
This is the exact problem.

#### pka

##### Elite Member
Find the value of the variable and GH if H is between G and I.
GI = 5b + 2, HI = 4b – 5, HI = 3
My work:
4b - 5 = 3
5 + 3 = 8
4b = 8
b = 2
GI = 5b + 2, and since b = 2
GI = 5(2) + 2
GI = 12
I don’t know how to get to GH.
$$\displaystyle \underbrace {G\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_H}_9\underbrace {\_\_\_\_\_\_\_\_\_\_I}_3$$
The above is what I get from $$\displaystyle b=2,~GI=12,~\&~HI=3$$

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#### ksdhart2

##### Senior Member
Having seen Pka's illuminating post, I now feel a bit foolish for not seeing that before, but that's tempered by some frustration over inconsistent notation. Obviously, students who post here have little to no control over what notation their textbooks and/or class handouts use. Yet, it creates confusion when the notation differs from problem to problem.

At crux here is that the OP's notation apparently uses $$GI$$ to refer to the length of a line segment whereas I'm so used to seeing $$GI$$ refer to the line segment as a whole and $$\overline{GI}$$ refer to that line segment's length, that it never occurred to me to consider that someone else might use different notation.

##### Junior Member
$$\displaystyle \underbrace {G\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_H}_9\underbrace {\_\_\_\_\_\_\_\_\_\_I}_3$$
The above is what I get from $$\displaystyle b=2,~GI=12,~\&~HI=3$$
Can you please explain to me how you got GH.

#### pka

##### Elite Member
Can you please explain to me how you got GH.
$$\displaystyle 12-3=9$$ With this posting you have convinced me that this material is really over your head.

#### pka

##### Elite Member
Is there a drawing showing G, H & I are collinear (in one st. line)?
Mr. Khan the statement that point $$\displaystyle B$$ is between points $$\displaystyle A~\&~C$$ written $$\displaystyle A-B-C$$ means that $$\displaystyle A,~B,~\&~C$$ are colinear and $$\displaystyle AB+BC=AC$$.

##### Junior Member
$$\displaystyle 12-3=9$$ With this posting you have convinced me that this material is really over your head.
Well, the problem is that I never learned it properly because the content for measuring segments in my textbook (which is the only form of learning) was actually different from what I had to apply to the assessments in the online course.

#### pka

##### Elite Member
Well, the problem is that I never learned it properly because the content for measuring segments in my textbook (which is the only form of learning) was actually different from what I had to apply to the question assessments in the online course.
Well I feel sorry for you. I gave up teaching because of on-line courses. I was trained in the Moore-method of teaching.
I just could not figure out how to teach a graduate mathematics course online.