# Maths vectors question

The question is:

OACB is a parallelogram with OA=a and OB =b. The points M,S,N,T divide OB,BC,CA and AO in the ratio 1:4 respectively. The lines ST and MN intersect at the point D. Show that the lines MN and ST bisect one another.
(edited 6 years ago)
Original post by dont know it
The question is:

OACB is a parallelogram with OA=a and OB =b. The points M,S,N,T divide OB,BC,CA and AO in the ratio 1:4 respectively. The lines ST and MN intersect at the point D. Show that the lines MN and ST bisect one another.

Ideally, you want to show that $\overrightarrow{SD} = \frac{1}{2}\overrightarrow{ST}$ and $\overrightarrow{MD} = \frac{1}{2}\overrightarrow{MN}$.

So, begin by determining what the point D is in terms of $\mathbf{a} . \mathbf{b}$.
Original post by RDKGames
Ideally, you want to show that $\overrightarrow{SD} = \frac{1}{2}\overrightarrow{ST}$ and $\overrightarrow{MD} = \frac{1}{2}\overrightarrow{MN}$.

So, begin by determining what the point D is in terms of $\mathbf{a} . \mathbf{b}$.

I got SD= u(3/5a-b) and MD=y(3/5b+a). I don't really get what to do after this.
Original post by dont know it
I got SD= u(3/5a-b) and MD=y(3/5b+a). I don't really get what to do after this.

Then you can play around and choose any path you want between two known coordinates, and express it in terms of the two you got.

I.e. We know what the path $\overrightarrow{MS}$ is in terms of $\mathbf{a}, \mathbf{b}$. We can also express it in terms of $\overrightarrow{SD}, \overrightarrow{MD}$. Make the two expressions equal to one another, then we must have the scalar multiples of $\mathbf{a}, \mathbf{b}$ equal on both sides of the equation. This gives us two linear eqs. in $u,y$ to solve simultaneously.
Original post by RDKGames
Then you can play around and choose any path you want between two known coordinates, and express it in terms of the two you got.

I.e. We know what the path $\overrightarrow{MS}$ is in terms of $\mathbf{a}, \mathbf{b}$. We can also express it in terms of $\overrightarrow{SD}, \overrightarrow{MD}$. Make the two expressions equal to one another, then we must have the scalar multiples of $\mathbf{a}, \mathbf{b}$ equal on both sides of the equation. This gives us two linear eqs. in $u,y$ to solve simultaneously.

I don't understand the paths bit. I tried to equate MD and SD together but I got it wrong. I got y(3/5b+a)=u(3/5a-b). When i expanded it and equated, I got 3y=5u and 3u=5y. That's obviously wrong.. but I'm not sure where I went wrong.
Original post by dont know it
I don't understand the paths bit. I tried to equate MD and SD together but I got it wrong.

No wonder. They are not equal to each other!

What's the path $\overrightarrow{MS}$ in terms of $\mathbf{a}, \mathbf{b}$ ?

What's the path $\overrightarrow{MS}$ in terms of $\overrightarrow{SD}, \overrightarrow{MD}$ ?

Make the two equal to each other.
Original post by RDKGames
No wonder. They are not equal to each other!

What's the path $\overrightarrow{MS}$ in terms of $\mathbf{a}, \mathbf{b}$ ?

What's the path $\overrightarrow{MS}$ in terms of $\overrightarrow{SD}, \overrightarrow{MD}$ ?

Make the two equal to each other.

Oh right got it now. Thank you.
Where did you get this question from?
why cant i find two routes for MD and equate ?