Answer:
(0, 0), (4, 0), (4, -5)
Step-by-step explanation:
This one keeps the same lengths of the previous triangle. It also keeps the same proportions (or something) etc, etc. I solved this problem by graphing and seeing which one fit best.
The artistic crop isn't helpful; it cuts off some vertex names.
The circumcenter H is the meet of the perpendicular bisectors of the sides, helpfully drawn. We have right triangle ELH, right angle L, so
EH² = HL² + EL²
EL = √(EH² - HL²) = √(5.06²-2.74²) ≈ 4.2539393507665339
Since HL is a perpendicular bisector of EF, we get congruent segments FL=EL.
Answer: 4.25