The area (in square units) of the triangle fo | vedclass

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(C) The given pair of straight lines is $x^2-3xy+2y^2=0$.Factoring the equation: $x^2-2xy-xy+2y^2=0$ $\Rightarrow x(x-2y)-y(x-2y)=0$ $\Rightarrow (x-y

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$\frac{1}{12}$

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(C) The given pair of straight lines is $x^2-3xy+2y^2=0$.Factoring the equation: $x^2-2xy-xy+2y^2=0$ $\Rightarrow x(x-2y)-y(x-2y)=0$ $\Rightarrow (x-y)(x-2y)=0$.This represents two lines: $L_1: x-y=0$ and $L_2: x-2y=0$.The third line is $L_3: x+y+1=0$.To find the vertices of the triangle,we find the intersection points of these lines:$1$. Intersection of $L_1$ and $L_2$: Solving $x=y$ and $x=2y$ gives $(0,0)$.$2$. Intersection of $L_1$ and $L_3$: Solving $x=y$ and $x+y+1=0$ gives $2x+1=0 \Rightarrow x=-\frac{1}{2}, y=-\frac{1}{2}$. Point is $(-\frac{1}{2}, -\frac{1}{2})$.$3$. Intersection of $L_2$ and $L_3$: Solving $x=2y$ and $x+y+1=0$ gives $2y+y+1=0$ $\Rightarrow 3y=-1$ $\Rightarrow y=-\frac{1}{3}, x=-\frac{2}{3}$. Point is $(-\frac{2}{3}, -\frac{1}{3})$.The area of the triangle with vertices $(0,0)$,$(-\frac{1}{2}, -\frac{1}{2})$,and $(-\frac{2}{3}, -\frac{1}{3})$ is given by:Area $= \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$Area $= \frac{1}{2} |0(-\frac{1}{2} - (-\frac{1}{3})) + (-\frac{1}{2})(-\frac{1}{3} - 0) + (-\frac{2}{3})(0 - (-\frac{1}{2}))|$Area $= \frac{1}{2} |0 + \frac{1}{6} - \frac{1}{3}| = \frac{1}{2} |-\frac{1}{6}| = \frac{1}{12}$ square units.

The area (in square units) of the triangle formed by $x+y+1=0$ and the pair of straight lines $x^2-3xy+2y^2=0$ is

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