And the intersections of the x axis and the y axis are b and c respectively.

The coordinate of point A is (-2,0).

(1) Try to explain that △ABC is an isosceles triangle;

(2) Moving point M starts from point A and moves along point B on the X axis, while moving point N starts from point B and moves along BC line to point C at a speed of 1 unit length per second. When one of the moving points reaches the end point, they all stop moving. When the point moves for t seconds, the area of △MON is S.

① Find the functional relationship between S and T;

② When the point M moves on the line segment OB, is there a situation that s=4? If it exists, find the corresponding t value; If it does not exist, explain the reasons;

③ When △MON is a right triangle in the process of motion, find the value of t. ..

Solution:

(1) Replace y= with y=0.

-4/3X+4, x = 3, ∴ the coordinate of point B is (3,0);

Replace y= with x=0.

, get y=4,

The coordinate of point ∴c is (0,4).

In Rt△OBC, oc = 4, OB=3, ∴BC=5.

And a (-2,0), ∴AB=5, ∴AB=BC, ∴△ABC are isosceles triangles.

(2)∵AB=BC=5, so point M and point N start moving at the same time and stop moving at the same time.

The crossing point n is nd ⊥ the x axis is d.

Then ND=NB●sin∠OBC=

①

When 0 < t < 2 (as shown in Figure A)

OM=2-t，

∴s= 1/2

Medal of merit

●ND=

1/2(2 tons) ●4/5 tons

=-2/5t？ +4/5 tons

When 2 < t ≤ 5 (Figure B), OM=t-2,

∴s=

=

=

................................., eight.

(Note: the value range of t is written as 0≤t≤2 and 2≤t≤5, respectively, and no points will be deducted. )

②

In one case, s=4.

When s=4,

=4

The solution is t 1= 1+

,

t2= 1-

second

........................ 10.

③

When MN⊥x axis and △MON are right triangles,

MB=NB●COS∠MBN=

, and MB = 5-T.

∴

=5-t，

∴t=

................ 1 1 min

When point M and point N move to point B and point C respectively, △MON is a right triangle with t=5.

Therefore, when △MON is a right triangle, t=

Seconds or t=5 seconds

.............. 12 point