Question

The area of triangle formed by $x+y+1=0$ and $x^2-3xy+2y^2=0$ is

Answer 1

$x+y+1=0$ and $x^2-3xy+2y^2=0$

Consider $x^2-3xy+2y^2=0$

$x^2-xy-2xy+2y^2=0$

$x(x-y)-2y(x-y)=0$

$(x-2y)(x-y)=0$

$\Rightarrow x-2y=0$ and $x-y=0$

$x-2y=0$

$(-)x-y=0$

____________ $y=0$ and $x=0$

$-y=0$

____________

$1^{st}$ point $(x_1,y_1)=(0,0)$

$\Rightarrow x+y+1=0$ and $x-y=0$

$x+y+1=0$

$x-y=0$ $x=-\frac{1}{2}$ and $y=\frac{1}{2}$

____________

$2x+1=0$

____________

$2^{nd}$ point $(x_2,y_2)=(-\frac{1}{2},\frac{1}{2})$

$\Rightarrow x+y+1=0$ and $x-2y=0$

$2y+y+1=0$

$y=\frac{-1}{3}$ and $x=2y$

$x=2\left(\frac{-1}{3}\right)=-\frac{2}{3}$

$3^{rd}$ point $(x_3,y_3)=(-\frac{2}{3},-\frac{1}{3})$

Area of a triangle formed by three points

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

$=\frac{1}{2}\left|[0(\frac{1}{2}-(-\frac{1}{3}))+(-\frac{1}{2})(-\frac{1}{3}-0)+(-\frac{2}{3})(0-\frac{1}{2})]\right|$

$=\frac{1}{2}\left|\left[\right[0]+\frac{1}{6}+\frac{2}{6}]\right|$

$=\frac{1}{2}\times \frac{3}{6}=\frac{1}{4}$ units.