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Math final exam! Answer high scores in detail! I want it before 4: 30 pm!
(1) According to the meaning of the question, the coordinates of point A are (8,0) and the coordinates of point B are (1 1.4).

And OA=BC, so that coordinate of point c is C(3, 4),

Let the analytical formula of straight line l be y=kx,

Substitute the coordinates of point c into y=kx,

The solution is k = 4 3,

The analytical formula of ∴ straight line L is y = 43x.

So, the answer is: (3,4), y = 43x.

(2) according to the meaning of the question, OP=t, AQ = 2t. It is discussed in three situations:

① when 0 < t ≤ 52, as shown in figure 1, the coordinate of point m is (t, 43t).

If the intersection point C in D is CD⊥x axis, and the intersection point Q in E is QE⊥x axis, then △AEQ∽△ODC can be obtained.

∴AQ OC =AE OD =QE CD,

∴2t 5 =AE 3 =QE 4,

∴AE=6t 5,EQ=8 5 t,

∴ The coordinates of point Q are (8+6 5 t, 8 5 t),

∴PE=8+6 5 t-t=8+ 1 5 t,

∴S= 1 2? MP? PE= 1 2? 4 3 t? (8+ 1 5t)= 2 15 T2+ 16 3t,

(2) When 52 < t ≤ 3, as shown in Figure 2, the intersection point Q is qf ⊥ the x axis is f,

∫BQ = 2t-5,

∴of= 1 1-(2t-5)= 16-2t,

∴ The coordinate of point Q is (16-2t, 4),

∴PF= 16-2t-t= 16-3t,

∴S= 1 2? MP? PF= 1 2? 4 3 t? ( 16-3t)=-2t2+32 3 t,

③ When point Q intersects point M, 16-2t=t, and the solution is t = 163.

When 3 < t < 163, as shown in figure 3, MQ= 16-2t-t= 16-3t, MP = 4.

s = 1 ^ 2? MP? PF= 1 2? 4? ( 16-3t)=-6t+32,

(3)① When 0 < t ≤ 52, S = 215t2+163t = 215 (t+20) 2-1603,

∫a = 2 15 > 0, parabolic opening is upward, and when t=5 2, the maximum value is 856;

② when 52 < t ≤ 3, s =-2t2+323t =-2 (t-83) 2+1289.

∵ A =-2 < 0, parabolic opening downward.

∴ When t = 8 3, s has a maximum value, namely128 9.

③ when 3 < t < 163, S=-6t+32,

∫k =-6 < 0。

∴S decreases with the increase of temperature.

When t=3, S = 14. When t= 16 3, S = 0.

∴0