Optimization problem: given that a line passes through $(4,3)$ and it forms a triangle with x and y axis, find minimum area I am asked to evaluate the following problem:

Given that a line $s$ passes through $(4,3)$ and it forms a triangle with $x$ and $y$ axis, find the minimum area shaped by it and the positive $x$ and $y$ axis.

It's easy to say that the area is half the base times the height, and that the equation of the line is 
$$
y = m (x-4) +3
$$
but how can I find $m$ or even $y$ in terms only of $x$ so I can differentiate and equal it to zero, finding a local minimum?
 A: Hint find it's intercepts on $x,y$ axis then use determinants(area with three vertices known) to create an equation in $m$ name it $f(m)$ and then differentiate set it equal to $0$ get the corresponding value of M and then get values of $x,y$ and thus the area. 
A: Calculate the crossings with $y$ and $x$ axis by setting the other parameter to zero:
$y = m(0-4) +3 \rightarrow y = -4m +3$ and
$0 = m(x-4) + 3 \rightarrow x = -\frac{3}{m}+4$
As you noted, area is half height times base:
$A(m) = \frac{1}{2}(3 - 4m)(4 - \frac{3}{m}) = 12 -\frac{9}{2m}-8m$
Can you solve from here?
A: We can use some principles of geometry as  check or a shortcut. 
Suppose we have a line through $(4,3)$ that intersects both the x and y axis. The we have a right triangle where this line is the hypotenuse and the acute angles are complementary. Rotate the line clockwise, the length of the leg along the y axis increases, the length of the leg along the x axis decreases and vice versa. As you rotate the hypotenuse line, the acute angles experience equal and opposite changes to maintain their sum at $\frac{\pi}{2}$ radians. 
Let $r1$ be the distance along the hypotenuse to the vertex on the $x$ axis, and let $r2$ be the distance along the hypotenuse to the $y$ axis.
A line a distance $r$ away sweeps out an area of $\frac{1}{2}r^2d\theta$ for a small turn $d\theta$. At small angular displacements, this approximates the area change of the area near the corresponding vertex of the triangle. 
We know the area changes near the the acute angles have opposite signs, so $$dA\approx\frac{1}{2}(r_2^2-r_1^2)d\theta$$
We want $\frac{dA}{d\theta}=0$ which only happens when $r1= r2$.
$(4,3)$ is thus the midpoint of the hypotenuse. In a right triangle, a circle centered at the midpoint of the hypotenuse having radius equal to the the length of the median from the right angle will intersect the all three of the triangle's vertices by Thales' Theorem. 
The midpoint of the hypotenuse is 5 units away from the origin. So the vertices on the x and y axes must also be 5 units away from the midpoint. 
So we have $$(x-4)^2+(y-3)^2=25$$
Holding $x=0$, then $y=0$ or $6$.
Holding $y=0$, then $x=0$ or $8$.
We can throw out $(0,0)$ so that leaves $(0,6)$ and $(8,0)$.
So $m=\frac{\delta y}{\delta x}=\frac{6}{-8}=\frac{-3}{4}$
