Given the points A(0,4), B(3,0), determine the coordinates of the point C that lies on the line x+y=0.The area of triangle ABC is 5 square units.

We'll determine the area of the triangle ABC , calculating
the determinant formed from the coordinates of the given
points.

The coordinates of C are :
C(a,b)

                | 0   4   1
|

A = (1/2)*| 3   0   1
|

                | a   b   1
|

We'll compute the
determinant:

A = (1/2)*(0*0*1 + 3*b*1 + 4*1*a - a*0*1 -
b*1*0 - 3*4*1)

A = (1/2)*(3b + 4a -
12)

But , from enunciation, the area of triangle ABC is: A
= 5 square units.

5 = (1/2)*(3b + 4a -
12)

3b + 4a - 12 = 10

We'll
add 12 both sides:

4a + 3b = 22
(1)

We know, from enunciation, that C is located on the
line x + y = 0.

a + b = 0
(2)

We'll subtract 3*(2) from
(1):

4a + 3b - 3a - 3b =
22

We'll combine and eliminate like
terms:

a = 22

From (2), a =
-b

b = -22

The
coordinates of the point C are: C(22 ,
-22).

Note: The line x + y = 0 represent the
bisectrix of the 2nd and 4th quadrants.