Question

if area of a triangle ABC formed by A(x,y) B(1,2) C(2,1) is 6 sq.units then prove that x+y=15 or x+y+9=0

Answer 1

Answer:

x+y=15 or x+y+9=0

Step-by-step explanation:

Formula used:

Area of a triangle having vertices $(x_1,y_1),\:(x_2,y_2)\:and\:(x_3,y_3)$ is

$|\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]|$

Given points are A(x,y) B(1,2) C(2,1)

Area of ΔABC = 6 sq. units

$|\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]|=6\\\\|\frac{1}{2}[x(2-1)+1(1-y)+2(y-2)]|=6\\\\|\frac{1}{2}[x(1)+1-y+2y-4]|=6\\\\|\frac{1}{2}[x+y-3]|=6\\\\\frac{1}{2}|[x+y-3]|=6\\\\|[x+y-3]|=12$

This implies

x+y-3=12 or x+y-3=-12

x+y=15 or x+y+9=0