Question

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

• A. $\frac{2}{\sqrt{3}}$
• B. $\frac{\sqrt{5}}{2}$
• C. $5\sqrt{2}$
• D. $\frac{1}{2\sqrt{5}}$

Answer 1

The correct option is D $\frac{1}{2\sqrt{5}}$

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

Substituting $x=-1-y$ in $x^2-3xy+y^2$, we get $(-1-y{)}^2-3y(-1-y)+y^2=0$

i.e. $y^2+2y+1+3y+3y^2+y^2=0$

$5y^2+5y+1=0$

$\Rightarrow y=\frac{-5\pm \sqrt{25-20}}{10}=\frac{-5\pm \sqrt{5}}{10}$

The intersection points would thus look like $(\frac{-5+\sqrt{5}}{10},\frac{-5-\sqrt{5}}{10})$ and $(\frac{-5-\sqrt{5}}{10},\frac{-5+\sqrt{5}}{10})$

The distance between these points gives the base of the triangle, which equals $\sqrt{\frac{1}{5}+\frac{1}{5}}=\sqrt{\frac{2}{5}}$

The length of perpendicular from origin (since it is the intersection of the given pair of straight lines) unto $x+y+1$ is equal to $\frac{1}{\sqrt{2}}$

Area $=\frac{1}{2}\times \frac{1}{\sqrt{2}}\times \sqrt{\frac{2}{5}}=\frac{1}{2\sqrt{5}}$