Vikas TU
Last Activity: 8 Years ago
Let the line be y = mx + c
passing through 3,4
the eqn. becomes ------>
4 = 3m +c-----------------------(A)
Area of triangle = -0.5 x c/m x c
= -0.5 x c^2/m
From eq. (A) substitute here,
f(c) let = 1.5c^2/(c-4)
form maximum f'(c) = 0
(2c(c-4) – c^2))/(c-4)^2 = 0
c^2 – 8c =0
c(c-8) =0
c = 0 or c =8
at c=8
Area of triangle = 24.sq. units.
