The area (in sq. units) of the triangle forme | vedclass

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(D) The given pair of lines is $x^2-3xy+y^2=0$. This represents two lines passing through the origin $(0,0)$.Let the lines be $y=m_1x$ and $y=m_2x$. T

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

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(D) The given pair of lines is $x^2-3xy+y^2=0$. This represents two lines passing through the origin $(0,0)$.Let the lines be $y=m_1x$ and $y=m_2x$. Then $m_1+m_2=3$ and $m_1m_2=1$.The area of the triangle formed by the lines $ax^2+2hxy+by^2=0$ and $lx+my+n=0$ is given by the formula:$	ext{Area} = \frac{n^2\sqrt{h^2-ab}}{|am^2-2hlm+bl^2|}$.Here,$a=1, h=-\frac{3}{2}, b=1, l=1, m=1, n=1$.Substituting these values:$	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}}$.

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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