There might be an easier way but here's one:
The area of the triangle is :
A = 0.5(x0)(y0), where x0 and y0 are the x and y intercepts. We want to minimise this.
The equation for a line is y = mx + y0, where m is the gradient and y0 is the y intercept.
Using (4,3) coordinate, we get 4 = 3m + y0, or y0 = 4 - 3m
now we need to find the x intercept in terms of m. This happens when y = 0:
0 = m(x0) + y0 = m(x0) + (4 - 3m)
Rearranging gives: x0 = 3 - 4/m
Now we can work out the area: A = 0.5 (x0) (y0) = 0.5 (3-4/m) (4-3m)
A = 0.5 (12 - 9m - 16/m + 12) = 0.5 (24 - 9m - 16/m)
To minimise the area, we can differentiate it with respect to m and let that equal 0:
dA/dm = d[ 0.5 ( 24 - 9m - 16/m) ] = 0
-9 + 32/m^2 = 0
m = +/- sqrt(32/9)
Because the line intercepts both x and y axes in the positive section, the gradient, m, has to be negative.
Substituting to find the y intercept to complete the equation:
c = 4 - 3m = 4 + 3(sqrt(32/9))
So the final answer is:
y = 4 + 3(sqrt(32/9)) - sqrt(32\9)x
or
y = 9.657 - 1.886x