1788–89 United States presidential election in Georgia
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Elections in Georgia |
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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.
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
- ^ Jensen & Becker 1976, p. xxix.
Works cited
- 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.
Categories:
- Articles with short description
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- Use mdy dates from September 2023
- Elections using electoral votes
- 1788–1789 United States presidential election by state
- United States presidential elections in Georgia (U.S. state)
- 1789 Georgia (U.S. state) elections
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