Question

Find the area of the triangle formed by the lines
$a{x}^2+2hxy+b{y}^2$ and $lx+my+n=0$

Answer 1

Let the lines represented by the given pair be $y=m_{ 1 }
x$and$y=m_{ 2 }x$, so that
$m_1+m_2=-2h/b$ and $m_1{m}_2=a/b$
These lines will meet the third line in points
$\left( \dfrac { -n }{ l+{ mm }_{ 1 } } ,\dfrac { { -nm }
_{ 1 } }{ l+{ mm }_{ 1 } } \right)$and$ \left( \dfrac
{ -n }{ l+{ mm }_{ 2 } } ,\dfrac { { -nm }_{ 2 } }{ l+{ mm
}_{ 2 } } \right)$
which will be two vertices of the triangle and the third
vertex is clearly $(0,0);$ hence the area of the
triangle when one vertex is at the origin is
$\dfrac{ 1 }{ 2 }(x_{ 1 }y_{ 2 }-x_{ 2 }y_{ 1 })=\dfrac{ 1
}{ 2 }\dfrac{ n^2{(m_{ 1 }-m_{ 2 })^2 } }{ (l+mm_{ 1 })(l
+mm_{ 2 }) }$
$=\dfrac{ 1 }{ 2 }n^{ 2 }\dfrac{ [(m_{ 1 }+m_{ 2 })^{ 2
}-4m_{ 1 }m_{ 2 }]^{ 1/2 } }{ l^{ 2 }+lm(m_{ 1 }+m_{ 2
})+m^{ 2 }.m_{ 1 }m_{ 2 } }$
$=\dfrac{ n^{ 2 }\surd (h^{ 2 }-ab) }{ am^{ 2 }-2hlm+bl^{
2 } }$.