1788–89 United States presidential election in Georgia

The 1788–89 United States presidential election in Georgia took place on January 7, 1789, as part of the 1788–89 United States presidential election. The state legislature chose 5 representatives, or electors to the Electoral College, who voted for President and Vice President.

1788–89 United States presidential election in Georgia

January 7, 1789 1792 →
 
Nominee George Washington John Milton James Armstrong
Party Independent Independent Independent
Home state Virginia Georgia Georgia
Electoral vote 5 2 1
Percentage 100.00%

 
Nominee Edward Telfair Benjamin Lincoln
Party Independent Independent
Home state Georgia Massachusetts
Electoral vote 1 1

President before election

Office established

Elected President

George Washington
Independent

George Handley, John King, George Walton, Henry Osborne, and John Milton served as electors. The electors cast five votes for George Washington, two for Milton, one for James Armstrong, one for Edward Telfair, and one for Benjamin Lincoln.[1]

References

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  1. ^ Jensen & Becker 1976, p. xxix.

Works cited

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  • Jensen, Merrill; Becker, Robert, eds. (1976). The First Federal Elections 1788-1790: Congress, South Carolina, Pennsylvania, Massachusetts, New Hampshire. Vol. 1. University of Wisconsin Press. ISBN 0299066908.