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.

Expert Answers

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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).

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