Given A (1,2), B (-3,1), then vector AB translates into vector (-1,2). Given A (1,2), B (-3,1), then vector AB is translated into vector (-1,2) and the vector coordinates are obtained.

A vector whose coordinates remain the same no matter how it is translated.
Therefore, the coordinates after translation are still (-3-1,1-2)=(-4,-1).(Only the coordinates of points A and B change)

A vector, no matter how it translates, its coordinates remain the same.
Therefore, the coordinates after translation are still (-3-1,1-2)=(-4,-1).(Only the coordinates of points A and B change)

In the parallelogram ABCD, the point M is the midpoint of AB, the point N is on BD and BN=1/3BD, prove that the three points M, N, C are collinear 0 Vector if possible In the parallelogram ABCD, the point M is the midpoint of AB, the point N is on BD and BN=1/3BD, prove that the three points M, N, C are collinear 0 Vectors if you can

∠ABD=∠BDC, MB/CD=BN/DN=1/2,△BMN CDN
Then ∠BMN=∠NCD, and MN/NC=1/2
It can be seen that M, N and C are collinear

In parallelogram ABCD, M is the middle point of AB, N is the upper point of BD, BN=1/3BD is proved: MNC three points are collinear, the process of finding the body is not In parallelogram ABCD, M is the middle point of AB, N is the last point of BD, BN=1/3BD is proved: MNC three points are collinear, the process of finding the body is not In the parallelogram ABCD, M is the middle point of AB, N is the last point of BD, BN=1/3BD is proved: MNC three points are collinear, the process of finding the body is not

Let the AB vector be a
AD vector is b
So AC=AB+AD=a+b
BD=BA+AD=b-a
Vector MN=MB+BN=AB/2+BD/3=a/2+(b-a)/3=a/6+b/3
Vector MC=MA+AC=-AB/2+AC=-a/2+a+b=a/2+b
So MC=3MN
So MC‖MN
And because the M.C.M.N. passed the M point,
So MNC three points collinear.

In the parallelogram ABCD, the point M is the midpoint of AB, and CM and BD intersect at the point N. If vector BN=λ vector BD, the value of λ is obtained. In the parallelogram ABCD, point M is the midpoint of AB, and CM and BD intersect at point N. If vector BN=λ vector BD, the value of λ is obtained.

∵△ BNM DNC
And the similarity ratio is 1/2,
Vector BN = vector BD/3
∴ λ = 1/3 .

In the parallelogram ABCD, the point M is the midpoint of AB, and CM and BD intersect at the point N. If the vector BN=λ vector BD, the value vector method of λ is obtained. Vector solution to do this kind of problem ideas.

BN=λBD=λ(BA+AD)=λBA AD
BN=1/2BA+MN=1/2BA+kMC=1/2BA+k (1/2BA+AD+AB)=1/2BA+kAD+k/2AB=(1-k)/2BA+kAD
Corresponding equality gives two equations
λ=(1-K)/2λ=k
λ=1/3

In parallelogram ABCD, vector AD=vector a vector AB=vector b M is the midpoint N of AB On DB, DN=2NB Verify MNC Three-Point Collinear

MN=MB+BN=1/2AB+1/3BD=1/2AB+1/3(BA+AD)=1/6b+1/3a
CN=CB+BN=-AD+1/3BD=-AD+1/3(BA+AD)=-2/3a+1/3b
So MN=-1/2CN
So MNC three-point collinear

Given A (1,2), B (-3,1), then vector AB translates into vector (-1,2). Given A (1,2), B (-3,1), then vector AB is translated into vector (-1,2) and the vector coordinates are obtained.