Spain’s Garbiñe Muguruza has defeated Swiss Timea Bacsinszky in straight sets 7-6(5) 6-2. It was a tight battle, and it was hard to determine who was leading, particularly in the first set, as they would both fight back hard if they were down. In the end, the world no. 18, lifted her game, to pull of the win.
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Photo: MELBOURNE, AUSTRALIA – JANUARY 19: Garbine Muguruza of Spain plays a forehand in her third round match against Timea Bacsinszky of Switzerland during day six of the 2019 Australian Open at Melbourne Park on January 19, 2019 in Melbourne, Australia. (Photo by Darrian Traynor/Getty Images)
In the first set, Bacsinszky broke in the opening game, but Muguruza instantly broke back. Bacsinszky then broke again in the fifth game, and was able to solidify the break to have a 4-2 lead. However, Muguruza broke back to put the games back on serve at 4-4. Muguruza had a chance to close out the set on the Swiss’ serve, but a backhand winner from Bacsinszky saved the set point and then a shot into the net by Muguruza allowed Bacsinszky to hold and take the set to a tie-break. Muguruza took the lead, and whilst the world no. 192 was able to catch up a little, it wasn’t enough, and Muguruza won the tie-break 7-5.
The second set had a similar start to the first, but this time, it was Muguruza who broke in the opening game, and Bacsinzky immediately broke back to level 1-1. In the fifth game again, Muguruza broke to lead 3-2 then consolidated. Muguruza found her form to break 5-2 and serve for the match. A ball sent long by Bacsinszky brought up three match points for Muguruza. It took her the third match point to close it out, but she was able to do so.
In the fourth round, Muguruza will play the winner of the match between Camila Giorgi and Karolina Pliskova.