# 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.

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We have two vertexes of the triangle as A(0,4) and B(3,0). Let the third vertex be C(m, n).

As C lies on x + y = 0, m + n = 0

=> m = -n

The line joining AB has the equation: y - 0 = [( 4 - 0)/ ( 0 - 3)](x - 3)

=> y = (-4/3)(x - 3)

=> 3y = -4x + 12

=> 4x + 3y -12 = 0.

The distance between the points A and B is sqrt(4^2 + 3^2) = 5.

The area or the triangle is given as 5.

Area = (1/2)*base*height

=> 5 = (1/2)*5*height

=> height = 2

So the distance of the point (m, n) from the line 4x + 3y -12 = 0 is 2.

=> | 4*m + 3*(-m) - 12|/ sqrt ( 3^2 + 4^2) = 2

=> (m - 12)/5 = 2

=> m - 12 = 10

=> m = 22

n = -m = -22

**Therefore the point C is (22, -22).**

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.