How would you work this out

Point P has position vector -7i+13j and point Q has position vector 13i-2j
Point R lies on the straight line joining P to Q and such that
PR:PQ = 3:5

Find a) how far R is from Q
b) the position vector of R
c) how far R is from the origin

Thanks

a)
distance PQ = sqrt ( (13- -7)^2 + (-2-13)^2) = 5
Distance PR = 3/5 * distance PQ = 3
Since R lies on PQ, QR = 2

b) PR:QR = 3:2
Know this formula?
R = PR/(PR+QR) * Q + QR/(PR+QR) * P
Note that the single letters in the line above are vectors.
R = 3/5 * (13,-2) + 2/5 * (-7,13) = (25/5,20/5) = (5,4)

c) Distance formula gives sqrt (5^2 + 4^2) = sqrt(41)