Question 1: (It is relatively easy to find the coordinates of point P first and then the values of λ and μ by analytic geometry. But the order of the two small questions is to solve the values of λ and μ first, and then the coordinates of point P. So 1 is solved by pure plane geometry. )

Solution: As shown in the above figure, if point N satisfies vector AN=2 and vector NC, then n must be on the AC side (if the vectors are equal, the direction must be the same, the same below), and there is AN = 2NC；; If AM and BN are connected, the intersection point is point p.

( 1)

Make an alternating parallel line through point M, intersect BN at point T, and intersect AB at point S.

NC = 2MT with point m as the midpoint of BC; At the same time AN=2NC.

Get a =4NC.

MT is parallel to AN, and the triangle APN is similar to the triangle MPT.

So there is AP=4MP.

AP/AM = AP/(AP+MP)= 4MP/(4MP+MP)= 4/5 = 0.8

That is, AP=0.8AM.

So λ=0.8

Pass through point C, make parallel lines of MA, CR and BN, and extend the line at point R.

Bp takes m as the midpoint of BC = pr = pn+NR

At the same time, it is easy to prove that the triangle CRN is similar to the triangle APN.

PN=2NR comes from AN=2NC.

BP/BN =(PN+NR)/(BP+PN)=(2NR+NR)/(PN+NR+2NR)=(3NR)/(5NR)= 0.6

That is, BP = 0.6BN billion.

So μ=0.6

(2) Meaning: The coordinates of A, B and C are (2, -2), (5, 2) and (3, 0) respectively.

It is easy to calculate that the coordinates of point M are (1, 1) and the coordinates of point N are (-4/3, -2/3).

By the linear equation (y-y1)/(y2-y1) = (x-x1)/(x2-x1)

Omit this process and you get:

The linear equation passing through the AM point is: y+3x-4=0.

The equation of a straight line passing through BN is: 19y -8x+2=0.

By solving the above two equations, the coordinate of point P is (1.2,0.4).

The remaining two questions will be answered when you have time.