Question

If the area of triangle $ABC$ formed by $A(x,y),B(1,2)$ and $C(2,1)$ is $6$ square units, then prove that $x+y=15$.

Answer 1

Given: $A(x,y)\equiv (x_1,y_1),B(1,2)\equiv (x_2,y_2)$ and $C(2,1)\equiv (x_3,y_3)$

We know than area of triangle $=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2((y_3-y_1))+x_3(y_1-y_2\left)\right]$

Then area of triangle whose vertex is $A(x,y),B(1,2)$ and $C(2,1)$ is

Area of $\Delta ABC=\left[x\right(2-1)+1(1-y)+2(y-2\left)\right]$

$=\frac{1}{2}(x+1-y+2y-4)$

$=\frac{1}{2}(x+y-3)$

But given that that the area of triangle is $6$ sq unit

$\therefore \frac{1}{2}(x+y-3)=6$

$\Rightarrow x+y-3=12$

$\Rightarrow x+y=15$