lenght of the hypotenus as a function of x?

Question

A right triangle is formed in the first quadrant by the x and y axis through the point (3,2) writ ethe length L of the hypotenuse as a function of x

ok I should add that you are also given points (0,y) and (x,0) I am not trying to find the slope of the line, It cannot be found since multiple lines are capable of being formed I am looking for the relationship between the lenght of the hypotnuse and the lenght of the x axis

the right angle is at point 0,0

Answer 1

There is an infinite number of right triangles formed using the x and y axes as the 90 deg angle, and some hypotenuse going through (3, 2). On the other hand, the constraint of having to pass through (3, 2) means that any selected length on the x-axis ({x>3}) defines a length on the y-axis:

For any point on the x-axis {x>3}, X (X, 0), the slope of line L is -(2-0)/(X-3) and I will call that -S. The general equation for the line L is y = -Sx + Y, where Y is the intercept on the y-axis when x=0.

We calculate Y by replacing x in the general equation with X, our actual chosen point, because there we know, y = 0.

This means Y = SX [y = 0 = -Sx + Y]. and the general point y can be found from the equation y = -Sx + SX = S(X-x).

But more to the point, having picked X (the length on the x-axis at the point (X, 0) and shown that Y=SX (where S=2/(X-3)) which is the length on the y-axis at the intercept (0, Y), the equation can be written for the length of line L , just in terms of the chosen value X, as was asked for.

L=SQRT(X^2 +(SX)^2)

QED

I must have missed the further explanation and hope it's a misunderstanding that you are "also given points (0,y) and (x,0)", because that make the question less interesting. Sorry. If the statement of the problem includes the end points, then the need for line L to extend through (3, 2) is built in and therefore irrelevant. You will still need the slope, but it's nothing more than the given y (actually -y) over the given x. Then, let the slope = m = -y/x, and y = -mx (there is no "b", or offset, since the origin is (0, 0) by the statement of the problem. The length of line L (that is, the hypotenuse), which I symbolized earlier as L itself, is then given by the Pythagorean equation:

L = SQRT(x^2 + (-mx)^2) This has the same form as my QED answer, but the "m" value is much more directly obtained when (x, 0) and (0, y) are provided in the statement of the problem.
y-axis
(0, y) given
|^\
|...\
|.....\
|......\ <...............slope = -y/x [=m], so y = -mx
|........L (label and length) L = SQRT(x^2 + y^2) = SQRT(x^2 + (-mx)^2)
|..........\
|............\<-----------(3, 2) given
|_______\(x, 0) given____x-axis

Note: an earlier version of this had y/x = m and y = mx. The slope is negative for all L {x>3 and intersecting (3, 2)}, so y = -mx is correct. However, (mx)^2 = (-mx)^2, so the length of L is the same as the earlier version showed.

Answer 2

Let me see if I understand the premise correctly,

Right triangle, first quadrant
passes through (3,2)
Length of Hypotenuse as function of x

the equation of the Hypotenuse can be found using the fact that the point (3,2) is part of the line

y = 2/3 x is the equation

Recall L = sqrt(x^2+y^2)

Substituting for y as a function of x

L = sqrt(x^2 + (2/3)^2x^2)

or L = sqrt(x^2(1+4/9))

L = x/3sqrt(13)

Good luck

Answer 3

are you picturing this?

maybe you should draw it to make sure you see it

this is a triangle that starts at the 0,0 spot and goes over 3 then up 2, then back to 0,0

if this is not what you are describing then I didn't get it right from your question, if I did then consider this:

since the hypoteneuse line goes through 0,0 then the y intercept is zero, and since it goes horizontally 3 units, and up 2 units, the slope is (rize over run) 2/3

the equation of the line (y=mx+b) is therefore y=3/2x

the equation for the length (L) of this hypoteneuse is L^2=x^2+y^2

the line above gives us y in terms of x so we can substitue

L^2=x^2+(3/2x)^2

and this just simplify as much as needed

L=squrt(x^2+(3/2 x)^2)

Answer 4

You use the formula a^2+b^2=c^2
The x coordinate is base and the y coordinate is the lenght.
2^2+3^2=c^2
4+9=c^2 square root both sides
13=c^2
3.60555=c

Answer 5

L is a defined value and does not depend on x or y. L is radical(13) - darn, I can't do math anymore in my old age - sheesh.

Answer 6

http://i4.photobucket.com/albums/y147/nixie_zh/hypotenus-1.jpg

See diagram above.

It can be proved easily that
(y-2)/3 = 2/(x-2) ... ... y=2x/(x-3)
Therefore length of the hypotenus = sqr(y^2 + x^2) = sqr(4x^2/(x-3)^2 + x^2), simplify and you will get:

x*sqr(x^2-6x+13)/(x-3)