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