analytic Geometry

Let the segment ends be (a,0) and (0,b)
If the segment passes through first quadrant point (5,2),
then a>0 and b>0.
The triangle has vertices (0,0), (a,0) and (0,b).
It is a right triangle with leg lengths a and b, and area ab/2=36 --> ab=72 --> a=72/b.
The equation of a line passing through (a,0) and (0,b) can be written as
x/a+y/b=1, and has slope=-b/a.
From the equation above, multiplying both sides times ab, we get
xb+ya=ab
We can rewrite it as xb+ya=72, because we had found that ab=72.
Substituting the coordinates for point (5,2), we get
5b+2a=72.
Substituting 72/b for a in the equation above, we get
5b+144/b=72.
Multiplying both sides of the equal sign times b, we get
5b^2+144=72b 5b^2-72b+144=0
The solutions are b=12 and b=12/5.
1) From b=12, we get a=72/12=6.
a) The length of the segment between (a,0) and (0,b) is
sqrt(a^2+b^2)=sqrt(12^2+6^2=6sqrt(5)=about 13.42
b) The slope of tha line segment is -12/6=-2.
c) what does "length segment" mean? Is this the same question as part a) ?

2) From b=12/5 we get a=72(5/12)=360/12=30.
a) The length of the segment between (a,0) and (0,b) is
sqrt(a^2+b^2)=sqrt((12/5)^2+30^2=6sqrt(629)/5=about 13.10
b) The slope of the line segment is (-12/5)(1/30)=-2/25=-0.08
c) what does "length segment" mean? Is this the same question as part a) ?