Geometry proof: is it enough?

[Levido]

question:

23b)
My working:


(excuse the messy working, it’s not to focus of this question it’s just to show)

Answer (from book): AG and AL both have the same gradient [MATH]\frac{2b-q-v}{2a-p-u}[/MATH]
My answer checks out. However, is it enough to say that because the gradient between A and L is equal to the gradient between A and G, G lies on the line AL? Surely G could lie outside of line AL. I attach a diagram if I’m unclear.

 

[Deleted member 4993]

question:
View attachment 15770
23b)
My working:
View attachment 15772

(excuse the messy working, it’s not to focus of this question it’s just to show)

Answer (from book): AG and AL both have the same gradient [MATH]\frac{2b-q-v}{2a-p-u}[/MATH]
My answer checks out. However, is it enough to say that because the gradient between A and L is equal to the gradient between A and G, G lies on the line AL? Surely G could lie outside of line AL. I attach a diagram if I’m unclear.

View attachment 15773

I would say:

Lines AG and AL have a common point (A) and those lines have same slope (gradient) \(\displaystyle \ \ \to \ \ \ \) points A, G and L are collinear.​

Good work - but present it "neater" before submitting.

 

[Dr.Peterson]

My answer checks out. However, is it enough to say that because the gradient between A and L is equal to the gradient between A and G, G lies on the line AL? Surely G could lie outside of line AL. I attach a diagram if I’m unclear.

Are you perhaps confusing the terms "line" and "line segment"?

All three of your G's are on the line AL; only G₁ is on the line segment AL.

 

[Levido]

Good work - but present it "neater" before submitting.

My apologies, it wasn’t necessary anyway but it was lazy sorry.

Are you perhaps confusing the terms "line" and "line segment"?

To be quite frank I had no idea about that terminology. Have definitely learnt something today.
The next question makes much more sense now too. No help needed for that though.

Thanks for the help. I will definitely do more research online before I ask here from now on.