1788–89 United States presidential election in Georgia

Main article: 1788–89 United States presidential election

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

1. 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.