Question

Show that the area of the triangle formed by the lines $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ and $lx+my+n=0$ is $\left| \dfrac{{{n}^{2}}\sqrt{{{h}^{2}}-ab}}{a{{m}^{2}}-2hlm+b{{l}^{2}}} \right|$ .

Answer 1

Hint: First of all, we should assume two lines that form the pair of equations, and then find the relations by multiplying the equations of these lines and comparing them with the given equation. Then, using the equations of these straight lines, we can find the three points of the triangle. Now since we know the vertices of triangle, we can calculate its area by using the formula $Area=\dfrac{1}{2}\left| \begin{matrix}
   {{x}_{1}} & {{y}_{1}} & 1 \\
   {{x}_{2}} & {{y}_{2}} & 1 \\
   {{x}_{3}} & {{y}_{3}} & 1 \\
\end{matrix} \right|$ .

Complete step by step solution:
We are given with the equation of pair of straight lines $a{{x}^{2}}+2hxy+b{{y}^{2}}=0...\left( i \right)$
By analysing this equation, we can say that it contains two straight lines that intersect at the origin. Let the origin be O.
Since both of these lines pass through origin O, we can assume the lines to be ${{l}_{1}}x+{{m}_{1}}y=0$ and ${{l}_{2}}x+{{m}_{2}}y=0$ .
Hence, we can write
$\left( {{l}_{1}}x+{{m}_{1}}y \right)\left( {{l}_{2}}x+{{m}_{2}}y \right)=0$
$\Rightarrow {{l}_{1}}{{l}_{2}}{{x}^{2}}+\left( {{l}_{1}}{{m}_{2}}+{{l}_{2}}{{m}_{1}} \right)xy+{{m}_{1}}{{m}_{2}}{{y}^{2}}=0$
We can compare this equation with equation (i), to get
$\begin{align}
  & {{l}_{1}}{{l}_{2}}=a \\
 & {{l}_{1}}{{m}_{2}}+{{l}_{2}}{{m}_{1}}=2h \\
 & {{m}_{1}}{{m}_{2}}=b \\
\end{align}$
Let us draw an arbitrary figure for simplification.

To calculate the area, we need the coordinates of the vertices of the triangle. We already have the coordinates of the origin O (0,0).
To find A, we need to solve ${{l}_{1}}x+{{m}_{1}}y=0$ and $lx+my+n=0$ simultaneously. Let us use the cross-multiplication method to solve these two equations.
$\dfrac{x}{\left( {{m}_{1}}n-0 \right)}=\dfrac{y}{\left( 0-{{l}_{1}}n \right)}=\dfrac{1}{\left( {{l}_{1}}m-l{{m}_{1}} \right)}$
Hence, we have
$\begin{align}
  & x=\dfrac{{{m}_{1}}n}{{{l}_{1}}m-l{{m}_{1}}} \\
 & y=\dfrac{-{{l}_{1}}n}{{{l}_{1}}m-l{{m}_{1}}} \\
\end{align}$
Thus, we get the coordinates of point A as $\left( \dfrac{{{m}_{1}}n}{{{l}_{1}}m-l{{m}_{1}}},\dfrac{-{{l}_{1}}n}{{{l}_{1}}m-l{{m}_{1}}} \right)$ .
Similarly, to find B, we need to solve ${{l}_{2}}x+{{m}_{2}}y=0$ and $lx+my+n=0$ simultaneously. Let us use the cross-multiplication method to solve these two equations.
$\dfrac{x}{\left( {{m}_{2}}n-0 \right)}=\dfrac{y}{\left( 0-{{l}_{2}}n \right)}=\dfrac{1}{\left( {{l}_{2}}m-l{{m}_{2}} \right)}$
Hence, we have
$\begin{align}
  & x=\dfrac{{{m}_{2}}n}{{{l}_{2}}m-l{{m}_{2}}} \\
 & y=\dfrac{-{{l}_{2}}n}{{{l}_{2}}m-l{{m}_{2}}} \\
\end{align}$
Thus, we get the coordinates of point B as $\left( \dfrac{{{m}_{2}}n}{{{l}_{2}}m-l{{m}_{2}}},\dfrac{-{{l}_{2}}n}{{{l}_{2}}m-l{{m}_{2}}} \right)$ .
We know that the area of a triangle with coordinates $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\text{ and }\left( {{x}_{3}},{{y}_{3}} \right)$ is
$Area=\dfrac{1}{2}\left| \begin{matrix}
   {{x}_{1}} & {{y}_{1}} & 1 \\
   {{x}_{2}} & {{y}_{2}} & 1 \\
   {{x}_{3}} & {{y}_{3}} & 1 \\
\end{matrix} \right|$
Thus, for the triangle OAB, we have
$Area=\dfrac{1}{2}\left| \begin{matrix}
   0 & 0 & 1 \\
   \dfrac{{{m}_{1}}n}{{{l}_{1}}m-l{{m}_{1}}} & \dfrac{-{{l}_{1}}n}{{{l}_{1}}m-l{{m}_{1}}} & 1 \\
   \dfrac{{{m}_{2}}n}{{{l}_{2}}m-l{{m}_{2}}} & \dfrac{-{{l}_{2}}n}{{{l}_{2}}m-l{{m}_{2}}} & 1 \\
\end{matrix} \right|$
By evaluating the determinant, we get
$Area=\dfrac{1}{2}\left| 1\left\{ \left( \dfrac{{{m}_{1}}n}{{{l}_{1}}m-l{{m}_{1}}}\times \dfrac{-{{l}_{2}}n}{{{l}_{2}}m-l{{m}_{2}}} \right)-\left( \dfrac{{{m}_{2}}n}{{{l}_{2}}m-l{{m}_{2}}}\times \dfrac{-{{l}_{1}}n}{{{l}_{1}}m-l{{m}_{1}}} \right) \right\} \right|$
$\Rightarrow Area=\dfrac{1}{2}\left| \left( \dfrac{{{m}_{1}}n}{{{l}_{1}}m-l{{m}_{1}}}\times \dfrac{-{{l}_{2}}n}{{{l}_{2}}m-l{{m}_{2}}} \right)-\left( \dfrac{{{m}_{2}}n}{{{l}_{2}}m-l{{m}_{2}}}\times \dfrac{-{{l}_{1}}n}{{{l}_{1}}m-l{{m}_{1}}} \right) \right|$
On simplifying the above equation, we get
$Area=\dfrac{1}{2}\left| \dfrac{{{l}_{1}}{{m}_{2}}{{n}^{2}}-{{l}_{2}}{{m}_{1}}{{n}^{2}}}{\left( {{l}_{1}}m-l{{m}_{1}} \right)\left( {{l}_{2}}m-l{{m}_{2}} \right)} \right|$
\[\Rightarrow Area=\dfrac{1}{2}\left| \dfrac{{{n}^{2}}\left( {{l}_{1}}{{m}_{2}}-{{l}_{2}}{{m}_{1}} \right)}{{{l}_{1}}{{l}_{2}}{{m}^{2}}-{{l}_{1}}l{{m}_{2}}m-{{l}_{2}}l{{m}_{1}}m+{{l}^{2}}{{m}_{1}}{{m}_{2}}} \right|\]
We know that ${{\left( a-b \right)}^{2}}=\left( a+{{b}^{2}} \right)-4ab$ .
We can also write the above identity as $\left( a-b \right)=\sqrt{\left( a+{{b}^{2}} \right)-4ab}$
Using this identity in the equation for area, we get
\[\Rightarrow Area=\dfrac{1}{2}\left| \dfrac{{{n}^{2}}\sqrt{{{\left( {{l}_{1}}{{m}_{2}}+{{l}_{2}}{{m}_{1}} \right)}^{2}}-4{{l}_{1}}{{l}_{2}}{{m}_{1}}{{m}_{2}}}}{{{l}_{1}}{{l}_{2}}{{m}^{2}}-lm\left( {{l}_{1}}{{m}_{2}}+{{l}_{2}}{{m}_{1}} \right)+{{l}^{2}}{{m}_{1}}{{m}_{2}}} \right|\]
Now, substituting the values ${{l}_{1}}{{l}_{2}}=a,\text{ }{{l}_{1}}{{m}_{2}}+{{l}_{2}}{{m}_{1}}=2h\text{ and }{{m}_{1}}{{m}_{2}}=b$ , we get
\[Area=\dfrac{1}{2}\left| \dfrac{{{n}^{2}}\sqrt{{{\left( 2h \right)}^{2}}-4ab}}{a{{m}^{2}}-lm\left( 2h \right)+b{{l}^{2}}} \right|\]
\[\Rightarrow Area=\dfrac{1}{2}\left| \dfrac{2{{n}^{2}}\sqrt{{{h}^{2}}-ab}}{a{{m}^{2}}-2hlm+b{{l}^{2}}} \right|\]
Hence, we get \[Area=\left| \dfrac{{{n}^{2}}\sqrt{{{h}^{2}}-ab}}{a{{m}^{2}}-2hlm+b{{l}^{2}}} \right|\] .
Hence, proved.

Note: We must note that the figure is just schematic, and not to scale. While calculating the area, we must not forget about the modulus sign. This problem involves a lot of calculation, so we must write each step carefully and avoid mistakes. We can use any method to solve the pair of equations while finding the points A and B.