SOLUTION: A right triangle is formed with vertices at the points (7,-2), (19,7), and (-5,14). What percent of the triangle's area lies in the fourth quadrant?

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Question 1062658: A right triangle is formed with vertices at the points (7,-2), (19,7), and (-5,14). What percent of the triangle's area lies in the fourth quadrant?
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!


The part of triangle ABC that lies in the fourth quadrant
is the triangle DAE.

You find the area of triangle ABC by the determinant:



The area of triangle ABC is 300.

You must now find the area of triangle DAE

But to do that you must find the coordinates of D and E:

You find the equation of line CA by using the slope formula: 



and the point-slope formula 



and after doing that and simplifying, you get the equation 
of line CA, which is,
 
4x + 3y = 22 

Then point D is the x-intercept of line CA, so substitute 0 for y
and solve for x and get the coordinates of D as

 

Now exactly the same way, you'll find the equation of the line BE
as 

3x - 4y = 29

Then point E is the x-intercept of line BA, so substitute 0 for y
and solve for x and get the coordinates of E as



Then use the matrix method again to find the area of triangle DAE:



Finally you need to find what percent 25/3 is of 300.

So you divide 



So the area of triangle DAE is 1/36th of the area of triangle ABC.

To find out what percent that is, we multiply by 100%



Edwin
Answer by ikleyn(52778)   (Show Source): You can put this solution on YOUR website!
.
Be careful with the solution by Edwin.

The area of the triangle is ONE HALF of the determinant Edwin wrote.

See the lesson
https://www.algebra.com/algebra/homework/Matrices-and-determiminant/Determinant-of-a-2x2-matrix-and-the-area-of-a-parallelogram-and-a-triangle.lesson

https://www.algebra.com/algebra/homework/Matrices-and-determiminant/Determinant-of-a-2x2-matrix-and-the-area-of-a-parallelogram-and-a-triangle.lesson

in this site.

OR see these links

https://people.richland.edu/james/lecture/m116/matrices/applications.html>https://people.richland.edu/james/lecture/m116/matrices/applications.html

https://people.richland.edu/james/lecture/m116/matrices/applications.html



http://demonstrations.wolfram.com/TheAreaOfATriangleUsingADeterminant/

http://mathforum.org/library/drmath/view/55063.html


So, the general path of the Edwin solution is correct, but intermediate numbers must be revised.



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