Question

If an equation $hxy+gx+fy+c=0$ represents a pair of lines, then:
(a) $fg=ch$
(b) $gh=cf$
(c) $fh=cg$
(d) $hf=-cg$

Answer 1

We know that the general form of second degree equation given in the following form: $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ and the condition when this equation represents pair of straight lines is as follows: $\left| \begin{matrix} a & h & g \\\ h & b & f \\\ g & f & c \\\ \end{matrix} \right|=0$ so similarly we are going to use this relation in the pair of straight lines given in the above problem: $hxy+gx+fy+c=0$. Here, find the values of a, b, c, h, g and f then substitute these values in the determinant which we have shown above.

Complete step by step answer:
In the above problem, we have given the equation for the pair of straight lines which is shown below:
$hxy+gx+fy+c=0$
We know the general equation of second degree is as follows:
$a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$
And when the above equation will represent pair of straight lines then the following relation will happen:
$\left| \begin{matrix} a & h & g \\\ h & b & f \\\ g & f & c \\\ \end{matrix} \right|=0$
So, we are going to use the above relation in the equation given in the above problem. Now, comparing the given equation with the second degree equation given in the above problem we are going to find the values of a, b, c, h, g, f.
$hxy+gx+fy+c=0$
$a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$
On comparing we got the following values:
$a=0,b=0$
Because there are no coefficients of ${{x}^{2}}\And {{y}^{2}}$.
As the “h” given in the above problem is half of that in the second degree equation. Similarly, g and f are also half of g and f respectively.
So, we are going to put $a\And b$ as 0, c will be remain c and h, g, f becomes $\dfrac{h}{2},\dfrac{g}{2},\dfrac{f}{2}$. Now, substituting these values in the above determinant and we get,
$\left| \begin{matrix} 0 & \dfrac{h}{2} & \dfrac{g}{2} \\\ \dfrac{h}{2} & 0 & \dfrac{f}{2} \\\ \dfrac{g}{2} & \dfrac{f}{2} & c \\\ \end{matrix} \right|=0$
To solve the above determinant, we are going to expand along row 1 and we get,
$\begin{aligned} & 0-\dfrac{h}{2}\left( \dfrac{h}{2}c-\dfrac{f}{2}\left( \dfrac{g}{2} \right) \right)+\dfrac{g}{2}\left( \dfrac{h}{2}\left( \dfrac{f}{2} \right)-0 \right)=0 \\\ & \Rightarrow -\dfrac{h}{2}\left( \dfrac{hc}{2}-\dfrac{fg}{4} \right)+\dfrac{g}{2}\left( \dfrac{h}{2} \right)\left( \dfrac{f}{2} \right)=0 \\\ & \Rightarrow -\dfrac{{{h}^{2}}c}{4}+\dfrac{fgh}{8}+\dfrac{fgh}{8}=0 \\\ & \Rightarrow -\dfrac{{{h}^{2}}c}{4}+\dfrac{2fgh}{8}=0 \\\ & \Rightarrow -\dfrac{{{h}^{2}}c}{4}+\dfrac{fgh}{4}=0 \\\ \end{aligned}$
Taking 4 as L.C.M in the above equation and we get,
$\begin{aligned} & \dfrac{-{{h}^{2}}c+fgh}{4}=0 \\\ & \Rightarrow -{{h}^{2}}c+fgh=0 \\\ \end{aligned}$
Taking h as common in the above problem and we get,
$\begin{aligned} & h\left( -hc+fg \right)=0 \\\ & \Rightarrow -hc+fg=0 \\\ & \Rightarrow hc=fg \\\ \end{aligned}$

So, the correct answer is “Option a”.

Note: From the above problem, we have learnt that if a second degree equation is given then what would be the condition for that equation to be a pair of straight lines. Also, you need to know how to solve the determinant otherwise you cannot solve this problem.