Quadrant by the x and y axes

[e^(i Pi)+1=0]

Homework Statement

A right triangle is formed in the first quadrant by the x and y axes and a line through the point (1,2). Find the vertices of the triangle such that its area is a minimum.

Homework Equations

$a=\frac{1}{2}xy$
$x^2+y^2=h^2$
$y-2=m(x-1)$
$p=x+y+h$
$p=x+m(x+1)+2+h$
$x^2+(mx-m+2)^2=h^2$

The Attempt at a Solution

The last several equations represent my attempt so far. I cannot figure out an equation to differentiate. I also tried trig functions, but that didn't help.

 

[NewtonianAlch]

Does this actually have a numerical answer? As far as I can see, you have two fixed points, the origin and the point (1,2), which means bringing the other point closer to the origin you decrease the area, and as long as it's not (0,0) it will remain a triangle.

 

[klondike]

You have to many variables that are not independent. You should be able to focus on one and only one, for example the slope of the the hypotenuse, then write area as a function of the slope and find a's extrema.

Homework Statement

A right triangle is formed in the first quadrant by the x and y axes and a line through the point (1,2). Find the vertices of the triangle such that its area is a minimum.

Homework Equations

$a=\frac{1}{2}xy$
$x^2+y^2=h^2$
$y-2=m(x-1)$
$p=x+y+h$
$p=x+m(x+1)+2+h$
$x^2+(mx-m+2)^2=h^2$

The Attempt at a Solution

The last several equations represent my attempt so far. I cannot figure out an equation to differentiate. I also tried trig functions, but that didn't help.

 

[Dick]

Concentrate on the second equation. Find the x and y intercepts in terms of m. Now use the first equation. Differentiate with respect to m to find the extermum.

 

[e^(i Pi)+1=0]

$x int: (m[x-1]+2, 0)$
$y int: (0, 2-m)$

how can I get rid of the x in the x intercept

edit: I can have $x=\frac{2a}{y}$, but doesn't seem to help.

am going to bed now, so will check in tomorrow, thanks for the assistance.

 

[SammyS]

$x int: (m[x-1]+2, 0)$
$y int: (0, 2-m)$

how can I get rid of the x in the x intercept

edit: I can have $x=\frac{2a}{y}$, but doesn't seem to help

The x-coordinate of x-intercept is equal to the base of the triangle.

The y-coordinate of y-intercept is equal to the altitude of the triangle.

How is the area of a triangle related to its base & altitude?

 

[Dick]

$x int: (m[x-1]+2, 0)$
$y int: (0, 2-m)$

how can I get rid of the x in the x intercept

edit: I can have $x=\frac{2a}{y}$, but doesn't seem to help.

am going to bed now, so will check in tomorrow, thanks for the assistance.

Put y=0 and solve for x to get the x-intercept. The result shouldn't contain an extra 'x'.

 

[e^(i Pi)+1=0]

Yes, I was really short on sleep when I wrote that. :zzz:

The equation is $\frac{1}{2}(2-m)(\frac{m-2}{m})=A$

I've got it now, thanks.