AP EAPCET 05-Jul-2022 Shift 2 Paper

Let positive vector of A,B,C and D are a,b,c and d respectively.
‌AB=b−a
‌AD=d−a
‌AC=AB+AD [ ∵ parallelogram law of vector ]
‌‌=b−a+d−a
‌c−a=b+d−a
‌c=b+d−a
‌c+a=b+d.....(i)
Position vector of M=‌

b+c

2

and Position vector of N=‌

c+d

2

AM+AN‌=(‌

b+c

2

−a)+(‌

c+d

2

−a)
‌=‌

b+c−2a+c+d−2a

2

‌=‌

b+d+x−4a

2

‌=‌

c+a+2c−4a

2

[ ∵ from Eq. (i) ]
‌=‌

3c−3a

2

=‌

3

2

(c−a)=‌

3

2

‌AC