Let the points of the triangle be # (x_1,y_1),( x_2, y_2),( x_3, y_3).#
Area of triangle is A = #((x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2))/2)#
Given # A = 4, (x_1,y_1) = (8,3), (x_2, y_2) = (5,4)#
Substituting we have the below Area equation:
#((8(4 – y_3) + 5(y_3 – 3) + x_3(3 – 4))/2 )= 4 #
#((8(4 – y_3) + 5(y_3 – 3) + x_3(3 – 4)) = 8#
#(32 – 8y_3) + (5y_3 – 15) + (-1x_3) = 8#
#17 – 3y_3 -x_3 = 8#
#– 3y_3 -x_3 = (8-17)#
#– 3y_3 -x_3 = -9#
#3y_3 + x_3 = 9# ----> Equation 1
Distance between points #(8,3), (5,4)# using distance formula is
#sqrt ( (8-5)^2+(3-4)^2)# = #sqrt ( 3^2+(-1)^2)# = #sqrt 10#
Distance between points #(x_3,y_3), (5,4)# using distance formula is
#sqrt ((x_3 -5)^2 + (y_3 - 4)^2)# = #sqrt 10#
Squaring both sides and subsituting #x_3 = 9 - 3y_3# from equation 1, we get a quadratic equation.
#(9-3y_3 - 5)^2 + (y_3 - 4 )^ 2 = 0#
#(4-3y_3)^2 + (y_3 - 4 )^ 2 = 0#
Factorizing this, we get # (y-1)(10y-22) = 0#
y = 1 or y = 2.2 . y=2.2 can be discarded. Hence, the third point has to be (6,1).
By calculating the distances for points #(8,3), (5,4) and (6,1)#, we get # sqrt 8# for the length of the base.