Submitted by maths123 on Sun, 05/31/2015 - 14:59

for 7ii - I used the method that since Q lies on both lines $P_1P_3$ and $ P_2P_4 $

$ q= p_1 + t(p_1 - p_3)$

$ q=p_2 + s(p_4 - p_2) $

hence equating the above two we find that

$ p_1 (1-t) + p_2 ( s-1) + tp_3 - sp_4 = 0 $

since there must exist real numbers t and s such that the above holds and as the sum of coefficients on the vectors is 0 it implies that

$ a_1 = 1-t$

$a_2 = s-1 $

$a_3 = t $

$a_4 = -s $

for 7iii for the first part can't you just say that, using the above values, $a_1 + a_3 = 1-t + t = 1 $ hence it is not equal to 0

also I do not understand the next expression given for $q$ - using the values for the constants above and subbing them into the first equation for q I find that $ q=a_1 p_1 + a_3 p_3 $ - why do I have to divide by $a_1 + a_3$ ?

## In part (iii) you need to

In part (iii) you need to consider

anynumbers $a_i$ which satisfy (*). You showed existence by explicitly constructing $a_i$ which satisfy (*), but unless you've shown uniqueness also, you can't use these values to demonstrate general results about $a_i$.## A quick example

Just to emphasise what's been said about uniqueness, it's obvious there can't be a unique solution: if $a_1,...a_4$ satisfy the equations, it's obvious that so do $ka_1,...ka_4$ for any constant $k$. This also hints at where the $a_1+a_3$ factor comes from: we need to divide by something to make the factors of $k$ disappear.