A line segment has its end on the coordinates axes and forms with then a triangle of area to 36 units. The segment passed through the point (5,2)

a). Compute the length of the line segment intercepted by the coordinates axes.

b). What is the slope of the line segment

c). compute the length segment

December 8, 2016 - 2:00pm

#1
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) ?