Expert: Paul Klarreich - 8/26/2010

Question
A right triangle is formed in the first quadrant by the x and
y axes (points (0,y) and (x,o)) and the line through the point
(3,2). Write the length of L the hypotenuse as a function of
x. Okay so I got part of it, but my teacher said something
about the distance formula and this ((x-3)^2+(y-2))^1/2

Answer
Questioner:   Pain
Category:  Calculus
Private:  no
 
Subject:  Trig/Precalc
Question:  A right triangle is formed in the first quadrant by the x and y axes (points (0,y) and (x,o)) and the line through the point (3,2). Write the length of L the hypotenuse as a function of x. Okay so I got part of it, but my teacher said something about the distance formula and this ((x-3)^2+(y-2))^1/2
 

let's make the points (0,b) and (a,0) -- we can fix it up later; it will make L be a function of 'a'.

Now the slope of the line is  -b/a, and it passes through (3,2)


y - 2 = -b/a (x - 3)

a(y - 2) = -b(x - 3)

ay - 2a = -bx + 3b

ay  = -bx + 3b + 2a
...................
y  = -b/a (x - 3) + 2

Now set  x = 0, y = b:

b  = -b/a (  - 3) + 2
 
ab  = 3b  + 2a

ab  - 3b  = 2a
         2a
b  = -------
       a - 3


Now we can determine the distance between  (a,0) and (0,b) using the D.F.

D = sqrt(a^2 + b^2)

D = sqrt(a^2 + (2a/(a-3))^2)

D = a sqrt(1 + (2/(a-3))^2)

D = a sqrt(1 + 4/((a-3)^2))

D = a/(a-3) sqrt(a^2 - 6a + 9 + 4)

D = a/(a-3) sqrt(a^2 - 6a +13)


NOTE: the web site seems to be mangling the characters. it writes 'percent 2B' in place of a plus sign. Sorry about that -- maybe you can fix it up.