# Minimizing Hypotenuse of a Right Triangle in Q1

Problem: A right triangle in the first quadrant has the coordinate axes as sides, and the hypotenuse passes through the point (1, 8). Find the vertices of the triangle such that the length of the hypotenuse is minimum.

My primary equation is: $$x^2 + y^2 = h^2$$ and I am minimizing $$h$$.

However, when selecting a secondary equation to relate $$x$$ and $$y$$ to express $$h$$ as a function of a single variable, two seemingly appropriate approaches yield different equations. Why is this?

Approach 1: Given three points, $$(x, 0), (1,8), (0,y)$$, we can find three equivalent expressions for slope $$m$$ of the line that describes the hypotenuse:

$$\text{slope} = \frac{y-8}{0-1} = \frac{0-8}{x-1} = \frac{y-0}{0-x}$$

Setting the first two expressions equal, the result is:

$$\frac{y-8}{0-1} = \frac{0-8}{x-1}$$

$$(y-8)(x-1) = 8$$

$$xy - 8x - y + 8 = 8$$

$$y(x-1) = 8x$$

$$y = \frac{8x}{x-1}$$

Approach 2: We solve for the equation of the line that describes the hypotenuse using a the point (1, 8) and the slope $$m = -\frac{y}{x}$$ (the last expression for $$m$$ above). Using the point-slope equation form of a line:

$$y - y_1 = m(x - x_1)$$ $$y - 8 = -\frac{y}{x}(x - 1)$$ $$xy - 8x = -y(x-1)$$ $$xy - 8x = -xy + y$$ $$2xy - y = 8x$$ $$y(2x-1) = 8x$$ $$y = \frac{8x}{2x-1}$$

Why do the results differ? What faulty assumption have I made? Thanks in advance.

The problem is that you are using the same variables meaning different things. In the second approach you are using $$(x,y)$$ on the line, while at the same time $$(x,0)$$ and $$(0,y)$$ are on the line. In particular, if $$(x,y)$$ is on the line, the slope is not $$-y/x$$