The area (in sq. units) of the triangle formed by the lines $x^2-3 x y+y^2=0$ and $x+y+1=0$, is
The area (in sq. units) of the triangle formed by the lines $x^2-3 x y+y^2=0$ and $x+y+1=0$, is
$\frac{2}{\sqrt{3}}$
$\frac{\sqrt{3}}{2}$
$5 \sqrt{2}$
$\frac{1}{2 \sqrt{5}}$
Solution
Given equations of line are
$
\begin{array}{r}
x^2-3 x y+y^2=0 \\
x+y+1=0
\end{array}
$
and
$
x+y+1=0
$
Let, $m_1$ and $m_2$ be the slope of the line
$
x^2-3 x y+y^2=2 \text {. }
$
$
\therefore \quad \begin{aligned}
\quad m_1+m_2 & =-\frac{\text { Coefficient of } x y}{\text { Coefficient of } y^2} \\
& =+\frac{3}{1}=3
\end{aligned}
$
and
$
\begin{aligned}
m_1 m_2 & =\frac{\text { Coefficient of } x^2}{\text { Coefficient of } y^2} \\
& =\frac{1}{1}=1
\end{aligned}
$
$
\text { Now, } \quad \begin{aligned}
m_1-m_2 & =\sqrt{\left(m_1+m_2\right)^2-4 m_1 m_2} \\
& =\sqrt{(3)^2-4 \times 1} \\
& =\sqrt{9-4}=\sqrt{5} \\
\Rightarrow \quad \quad \quad m_1-m_2 & =\sqrt{5}
\end{aligned}
$
On solving Eqs. (i) and (iii), we get and
$
\begin{aligned}
& m_1=\frac{3+\sqrt{5}}{2} \\
& m_2=\frac{3-\sqrt{5}}{2}
\end{aligned}
$
$\therefore$ Equation of lines will be
and
$
\begin{aligned}
& y=\frac{3+\sqrt{5}}{2} x \\
& y=\frac{3-\sqrt{5}}{2} x
\end{aligned}
$
and third line is given
$
x+y+1=0
$
$\therefore$ The points of intersection of these lines are $A(0,0), B\left(-\frac{2}{5+\sqrt{5}}, \frac{3+\sqrt{5}}{5+\sqrt{5}}\right)$, and
$
C\left(-\frac{2}{5-\sqrt{5}}, \frac{3-\sqrt{5}}{5-\sqrt{5}}\right) \text {. }
$
$\therefore$ Area of triangle
$
\begin{aligned}
& =\frac{1}{2}\left|\begin{array}{ccc}
0 & 0 & 1 \\
-\frac{2}{5+\sqrt{5}} & \frac{3+\sqrt{5}}{5+\sqrt{5}} & 1 \\
-\frac{2}{5-\sqrt{5}} & \frac{3-\sqrt{5}}{5-\sqrt{5}} & 1
\end{array}\right| \\
= & \frac{1}{2}\left[0+0+1\left(\frac{-2(3-\sqrt{5})}{5^2-(\sqrt{5})^2}+\frac{2(3+\sqrt{5})}{5^2-(\sqrt{5})^2}\right]\right. \\
= & \frac{1}{2}\left[\frac{-6+2 \sqrt{5}+6+2 \sqrt{5}}{25-5}\right] \\
= & \frac{1}{2}\left[\frac{4 \sqrt{5}}{20}\right]=\frac{1}{2 \sqrt{5}}
\end{aligned}
$