Fnd the area A of the triangle with the given the vertices

[Medtner]

(0, 0), (3, 5), (1, 8)

Find the slopes and equations for each line

(0,0) ----> (3,5) = 5/3x
(0,0)---->(1,8) = 8x
(1,8)---->(3,5) = -3/2x+ 19

Then I set up the integrals (on x)

Integral sign from 0 to 1 (8x-5/3x)dx + Integral sign from 1 to 3 [(-3/2x+19)-5/3x) dx

I got 117/4 as an answer and that's wrong. My algebra/arithmetic isn't wrong (i triple checked it) so it must have something to do with the set up. What's wrong with it?

 

[lewando]

For the line specified by (1,8) and (3,5), what algebra/arithmetic did you use to get a y-intercept of 19?

 

[lurflurf]

that should be

(0,0) ----> (3,5) = 5/3x
(0,0)---->(1,8) = 8x
(1,8)---->(3,5) = -3/2x+ 19/2[/color]

I suppose the directions asked for an integral it is not the best way otherwise

 

[lurflurf]

For the line specified by (1,8) and (3,5), what algebra/arithmetic did you use to get a y-intercept of 19?

3+8*2=19
then divide by 2

 

[Medtner]

Wait, 19/2?

y=mx+b
For the slope i got (-3/2)
used point (3,5)
5=-3/2(3)+b
10=-9
19=b

Did I do something wrong?

 

[lurflurf]

should be

Wait, 19/2?

y=mx+b
For the slope i got (-3/2)
used point (3,5)
5=-3/2(3)+b
10=-9+2b
19=2b
b=19/2[/color]

Did I do something wrong?

 

[Medtner]

Where did you get the 2b from?

 

[lurflurf]

when you multiply both sides by 2 also multiply b

 

[Nidum]

Have you been specifically told to use calculus for this problem ?

If not then the area of the triangle can be calculated directly from the vertex coordinates .

Try searching on ' finding area of a triangle using coordinate geometry '

and for general interest ' Shoelace formula '

If this is homework then it should really be in the PF homework section .