Find the equation of the line through the point (3,4) which cuts from the first quadrant a triangle of minimum area.

Answer 1

Draw a line through (3,4) cutting the x-axis at (x,0) and the y-axis at (0,y)

Area = (1/2(x)(y)

using slopes,
(y-4)/-3 = +4/(x-3)
xy - 3y - 4x + 12 = 12
y(x-3) = 4x
y = 4x/(x-3)

then
A = (1/2)x(4x)/x-3)
= 2x^2/(x-3)

dA/dx = [(x-3)(4x) - 2x^2]/(x-3)^2 = 0 for a max/min of A
4x^2 - 12x - 2x^2 = 0
2x^2 - 12x = 0
2x(x-6)= 0
x = 0 or x = 6, and y = 24/3 = 8

using (0,8) and (3,4)
slope = (8-4)/-3
= -4/3

line has equation
y = (-4/3)x + 8
slope of line =

Answer 2

" using slopes,
(y-4)/-3 = +4/(x-3)
xy - 3y - 4x + 12 = 12 "

can you explain that part

Answer 3

nvr mind
Anyways thanks Reiny

Answer 4

Find the gradient of a line joining R(4,8) & S(5,-2).