Answer:
Solution: As shown in the figure, the intersection point D is DF⊥BC of point F,
Bca = 90, AC=BC=2√2, the image of inverse proportional function y = 3/x (x > 0) intersects AB and BC at points D and E, ∴∠BAC =∠ABC = 45°, which can be set as E.
∴C(a,0),B(a,2√2),A(a-2√2,0),
∴ The analytical formula for finding straight line AB is: y = x+2 √ 2-a. 。
∫△BDE∽△BCA,
∴△BDE is also an isosceles right triangle,
∴DF=EF,
∴a-b=3/b-3/a, namely AB = 3.
Point d is on the straight line AB,
∴(3/b)=b+2√2-a, that is, 2A 2-(2 √ 2) A-3 = 0, and the solution is a=(3/2)√2.
The coordinates of point E are ((3/2)√2, √2).
So the answer is: ((3/2)√2, √2).