ABCD is a parallelogram M is the midpoint of CB
N is the midpoint of AB
DA = a
DC = c
a) find, in terms of a and c, the vector of MN
b) show that CA is parralel to MN

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Step-by-step explanation:

a) MN= 0.5(a-c)

b) CA=a-c

MN= 0.5(a-c)

MN=0.5xCA

(I got full marks for this answer)

Part a)

MN = 0.5a-0.5c

Part b)

MN=0.5a-0.5c

CA=a-c

CA is parallel to MN because 2×MN = CA.

Step-by-step explanation:

Part a = We know that ABCD is a parallelogram (stated), thus opposite sides must be equal. So given that DA = a, CB = a, DC = c, AB = C.

MN = MB+BN, so as M is the midpoint of CB then MB = 1/2 a and as we are going against AB (we are doing BN) then it will be -0.5c as N is the midpoint of AB.

Part B= More of a mathswatch error, where you have to quite literally put the precise answer. CA = CD+DA so -c (as we are going the opposite way to DC) + a.

CA = -c+a

Now CA is a multiple of MN. But Mathswatch prefers if you say that CA is parallel to MN because MN times 2 is equal to CA

tube or goog would be a great resource to with that question

the answer would be c:   ax + b = ax + c

step-by-step explanation: