calculus HELP URGENT
posted by Santha .
Find the equation of the line through the point (3,4) which cuts from the first quadrant a triangle of minimum area.

Draw a line through (3,4) cutting the xaxis at (x,0) and the yaxis at (0,y)
Area = (1/2(x)(y)
using slopes,
(y4)/3 = +4/(x3)
xy  3y  4x + 12 = 12
y(x3) = 4x
y = 4x/(x3)
then
A = (1/2)x(4x)/x3)
= 2x^2/(x3)
dA/dx = [(x3)(4x)  2x^2]/(x3)^2 = 0 for a max/min of A
4x^2  12x  2x^2 = 0
2x^2  12x = 0
2x(x6)= 0
x = 0 or x = 6, and y = 24/3 = 8
using (0,8) and (3,4)
slope = (84)/3
= 4/3
line has equation
y = (4/3)x + 8
slope of line = 
" using slopes,
(y4)/3 = +4/(x3)
xy  3y  4x + 12 = 12 "
can you explain that part 
nvr mind
Anyways thanks Reiny 
Find the gradient of a line joining R(4,8) & S(5,2).
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