Extend the line AB to C so that BC = 1 / 2Ab and D is the midpoint of AC. then extend AB to e reversely so that EA = ad. if AB = 6cm, calculate the length of AE, Use ∵

E——A——D——B——C
∵AC=AB+BC,BC=AB/2
∴AC=AB+AB/2=3AB/2
∵AB=6
∴AC=6×3/2=9
∵ D is the midpoint of AC
∴AD=AC/2=9/2=4.5
∵AE=AD
∴AE=4.5（cm）
Sorry, I'm not in at noon,

Let AB = 4. Extend AB to C so that BC = 12ab. D is the midpoint of AC, and extend AB to e reversely so that EA = ad. find the length of AE

As shown in the figure: ∵ line AB = 4cm, BC = 12ab, ∵ BC = 2cm, ∵ AC = 4 + 2 = 6cm, ∵ D is the midpoint of AC, ∵ ad = 3cm, ∵ EA = ad. ∵ AE = 3cm

Given a (- 2,4), B (3, - 1), C (- 3, - 4), let vector AB = a, vector BC = B, vector CA = C, (1) find 3A + B, (2) find the real number Mn satisfying a = MB + NC

(1)
a=(3,-1)-(-2,4)=(5,-5),b=(-6,-3),c=(1,8)
So 3A + B = (9, - 18)
(2)
If a = MB + NC, then (3, - 1) = m (- 6, - 3) + n (1,8) = (- 6m + N, - 3M + 8N)
So - 6m + n = 3
-3M + 8N = - 1, M = - 5 / 9, n = - 1 / 3

A (2,4) B (3, - 1) C (- 3, - 4). Let vector AB = a, vector BC = B, vector CA = C (1) find the coordinates of 3A + b-3c (2). Let B and C be a set of bases to represent a

(1) a=(1,-5),b=(-6,-3),c(5,8)∴3a+b-3c=(3,-15)+(-6,-3)+(-15,-24)=(-18,-42)(2) a=r1b+r2c∴(1,-5)=(-6r1,-3r1)+(5r2,8r2),∴1=-6r1+5r2,-5=-3r1+8r2∴r1=r2=-1a=-b-c

Extend the line AB to C so that BC = 1 / 2Ab and D is the midpoint of AC. then extend AB to e reversely so that EA = ad. if AB = 6cm, calculate the length of AE, Use ∵