It is known that both the image of inverse proportional function y= 12/x and the image of linear function y=kx-7 pass through point P(m, 2). There are two points A and B on the image of this linear function, which are perpendicular to the X-axis and intersect with the image of this inverse proportional function at C and D. If CD=AB, and the abscissas of A and B are 2 and 2+a respectively.

(1) Find the value of a (2) Find the perimeter of the quadrilateral ABDC.

Solution: (1). Point P(m, 2) is on the image of inverse proportional function y= 12/x, so there is 2= 12/m, that is, m =12/= 6; P (6 6,2) at a time.

The function y=kx-7, so there is 2=6k-7, that is, k = 9/6 = 3/2; So the equation of linear function is: y = (3/2) x-7;

According to the conditions, the coordinates of four points A, B, C and D are: A(2, -4), B(2+a, (3/2)(2+a)-7)=(2+a, 3a/2-4);

C(2，6)，D(2+a， 12/(2+a))； Equation from ∣AB∣=∣CD∣:

[(2+a)-2)？ +[(3a/2-4-(-4)]？ =[(2+a)-2]？ +[ 12/(2+a)-6]？

Simplify to get an A? +9a？ /4=a？ +36a？ /(2+a)？

(2+a)？ = 16, 2+A = 4, so a=2 or a=-6, it is discarded.

Namely A(2,-4); B(4，- 1)； C(2，6)； D(4，3).

(2)∣AB∣=√[(4-2)？ +(- 1+4)? ]=√ 13; ∣CD∣=√[(4-2)？ +(3-6)? ]=√ 13; ∣ac∣=4+6= 10；

∣bd∣= 1+3=4； Therefore, the perimeter of quadrilateral ABDC = 14+2√ 13.