Vectors...really really need help!?

Question

With respect to the origin O, the position vectors of points A, B and C are (2, 3), (4, -2) and (-3, -3) respectively.

(i) A point E (-2, p) is such that 2 |CE| = |AB|. Find the possible values of p.
Ans: - 5 1/2 or -1/2

(ii) Find the coordinates of the point Q if 2 CQ = BA + 1/2 AC.
Ans: (-5 1/4, -2)

Please help by showing workings...thanks

Answer 1

(i)
|CE|^2 = (-3 - (-2))^2 + ((-3) - p)^2
= 1 + 9 + p^2 + 6p
= p^2 + 6p + 10.

|AB|^2 = (2 - 4)^2 + (3 - (-2))^2
= 4 + 25
= 29

Therefore, as 4 CE^2 = AB^2:
4p^2 + 24p + 40 = 29
4p^2 + 24p + 11 = 0
(2p + 11)(2p + 1) = 0
p = -11/2 or - 1/2.

(ii)
BA = (2, 3) - (4, -2) = (-2, 5)
AC = (-3, -3) - (2, 3) = (-5, -6)

2CQ = (-2 , 5) + (1/2)(-5, -6)
= (-2 - 5/2, 5 - 3)
= (- 9/2, 2) ...(1)

If Q is (x, y), then:
2CQ = 2(x - (-3), y - (-3))
= (2x + 6, 2y + 6) ...(2)

Equating components from (1) and (2):
2x + 6 = -9/2
x = ( - 9/2 - 6) / 2 = -21/4,
and
2y + 6 = 2
y = (2 - 6) / 2 = - 2

Thus Q = (-21/4, - 2).

Answer 2

A) 2i+3j
B)4i-2j
C)-3i-3j
AB ) 2i-5j and IABI =sqrt(29)
CE ) i+( p+3)j and ICE I = sqrt(1+(p+3)^2)
2sqrt((1+(p+3)^2) = sqrt29
4 +4(p+3)^2 =29
4p^2+24p+11=0 p= (-24+-20)/8 so p= -1/2 and p=-5 1/2
Call Q) x i+y j
2[(x+3)i+(y+3)j] = -2i+5j -5/2i-3j
2x+6= -9/2 x= -21/4 = -5 1/4
2y+6=2 so y=-2