Question

Find the straight lines that are represented by the equation ${{x}^{2}}-7xy+12{{y}^{2}}=0$ and determine the angle between them.
A. $x-3y=0;x-4y=0;\theta ={{\cos }^{-1}}\left( \dfrac{13}{\sqrt{170}} \right)$
B. $x+3y=0;x+4y=0;\theta ={{\cos }^{-1}}\left( \dfrac{13}{\sqrt{170}} \right)$
C. $x-3y=0;x+4y=0;\theta ={{\cos }^{-1}}\left( \dfrac{13}{\sqrt{170}} \right)$
D. $x+3y=0;x-4y=0;\theta ={{\cos }^{-1}}\left( \dfrac{13}{\sqrt{170}} \right)$

Answer 1

Hint: In this question, we are to find one of the equations of a pair of lines represented by the given equation. For this, we need to know the general form of the equation of the pair of straight lines. So, we can compare them and get the value of the variable coefficients in order to submit them in the formula to find the required equation. The angle between those two lines is obtained by the $\cos \theta $ formula.



Formula Used:The equation of the pair of straight lines is
$H\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}=0$
This is called a homogenous equation of second degree in $x$ and $y$
And
$S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$
This is called a general equation of the second degree in $x$ and $y$.
If ${{h}^{2}}If ${{h}^{2}}=ab$, then $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents coincident lines.
If ${{h}^{2}}>ab$, then $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents two real and different lines that pass through the origin.
Thus, the equation $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents two lines. They are:
$ax+hy\pm y\sqrt{{{h}^{2}}-ab}=0$
The angle between these lines is
$\cos \theta =\dfrac{\left| a+b \right|}{\sqrt{{{(a-b)}^{2}}+4{{h}^{2}}}}$



Complete step by step solution:Given the equation of pair lines as
${{x}^{2}}-7xy+12{{y}^{2}}=0\text{ }...(1)$
But we have the homogenous equation of pair of lines as
$a{{x}^{2}}+2hxy+b{{y}^{2}}=0\text{ }...(2)$
Here, this equation represents two lines. They are:
$ax+hy\pm y\sqrt{{{h}^{2}}-ab}=0$
On comparing (1) and (2), we get
$a=1;b=12;h=\dfrac{-7}{2}$
Thus, the lines represented by the given equation are
$\begin{align}
  & ax+hy\pm y\sqrt{{{h}^{2}}-ab}=0 \\
 & \Rightarrow (1)x+(\dfrac{-7}{2})y\pm y\sqrt{\dfrac{49}{4}-(1)(12)}=0 \\
 & \Rightarrow 2x-7y\pm 2y\sqrt{\dfrac{49-48}{4}}=0 \\
 & \Rightarrow 2x-7y\pm y=0 \\
\end{align}$
Then, by separating their signs and simplifying them, we get
$\begin{align}
  & 2x-7y+y=0 \\
 & \Rightarrow 2x-6y=0 \\
 & \therefore x-3y=0 \\
\end{align}$
And
$\begin{align}
  & 2x-7y-y=0 \\
 & \Rightarrow 2x-8y=0 \\
 & \therefore x-4y=0 \\
\end{align}$
Thus, the lines of ${{x}^{2}}-7xy+12{{y}^{2}}=0$ are $x-3y=0;x-4y=0$.
Then, the angle between these two lines represented by ${{x}^{2}}-7xy+12{{y}^{2}}=0$ is
$\begin{align}
  & \cos \theta =\dfrac{\left| a+b \right|}{\sqrt{{{(a-b)}^{2}}+4{{h}^{2}}}} \\
 & \text{ }=\dfrac{\left| 1+12 \right|}{\sqrt{{{(1-12)}^{2}}+4{{\left( \dfrac{-7}{2} \right)}^{2}}}} \\
 & \text{ }=\dfrac{13}{\sqrt{121+49}} \\
 & \text{ }=\dfrac{13}{\sqrt{170}} \\
\end{align}$
$\therefore \theta ={{\cos }^{-1}}\left( \dfrac{13}{\sqrt{170}} \right)$



Option ‘A’ is correct



Note: Remember that, if we have the equation of pair of lines, then to know the lines by which it is formed, we can compare it with the general form or homogenous form of the lines to get the coefficients of the variables of the lines. So, by substituting them in the formulae, we get the required lines. For finding the angle between these lines, the combined equation is used in the angle formula.