Question

QP is (x y) = (3 -1) + r (14 10)
QR is (x y) = (3 -1) + s (17 -9)
PR is (x y) = (0 18) + t (5 -7)
where r, s, and t are scalars.

1. Use vector methods to find the coordinates of P, Q, and R.
2. Find vectors PQ and PR and evaluate PQ*PR
3. Find size of QPR
4. Find the area of triangle PQR

Answer 1

Your notation is slightly different from the one I am used to.

QP is a line with (x,y) = (3,-1) + r(14,10)
or
x = 3 + 14r
y = -1 + 10r

PR:
x = 0 + 5t
y = 18 - 7t

P is the intersection of QP and PR
so
3 + 14r = 0 + 5t ----> 14r - 5t = -3 **
-1 + 10r = 18 - 7t ---> 10r + 7t = 19 ***

** times 7 ----> 98r - 35t = -21
*** times 5 ---> 50r + 35t = 95
add them:
148r = 74
r = 1/2

then in
x = 3 + 14r -----> x = 3 + 7 = 10
y = -1 + 10r ----> y = -1 + 5 = 4
P is (10,4)

Find Q and R the same way.
(I bet Q is (3,-1) , how do you think I found that without any work ??)

#2, I will do vector PQ
vector PQ = [3-10, -1-4] or [-7,-5]

if PQ*PR is the 'dot' product, then you should have a method in your text or notes, it is one of the easiest calculations

(in my notation, we use [....] for vectors and (...) for points )

#3 I don't know what you mean by "size" of QPR, to me that would mean area.
In that case what is the difference between #3 and #4 ?