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(D) The area of the triangle formed by the pair of lines $ax^2+2hxy+by^2=0$ and the line $lx+my+n=0$ is given by the formula $	ext{Area} = \frac{n^2\

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$\frac{1}{2\sqrt{5}}$

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(D) The area of the triangle formed by the pair of lines $ax^2+2hxy+by^2=0$ and the line $lx+my+n=0$ is given by the formula $	ext{Area} = \frac{n^2\sqrt{h^2-ab}}{|am^2-2hlm+bl^2|}$.Here,the pair of lines is $x^2-3xy+y^2=0$,so $a=1, h=-\frac{3}{2}, b=1$.The third line is $x+y+1=0$,so $l=1, m=1, n=1$.Substituting these values into the formula:$	ext{Area} = \frac{1^2\sqrt{(-\frac{3}{2})^2-(1)(1)}}{|(1)(1)^2-2(-\frac{3}{2})(1)(1)+(1)(1)^2|}$$	ext{Area} = \frac{\sqrt{\frac{9}{4}-1}}{|1+3+1|} = \frac{\sqrt{\frac{5}{4}}}{5} = \frac{\frac{\sqrt{5}}{2}}{5} = \frac{\sqrt{5}}{10} = \frac{1}{2\sqrt{5}}$ sq. units.

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

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