Question

If the equation \[6{x^2} - 11xy - 10{y^2} - 19y + c = 0\] represents a pair of lines, find their equations. Also find the angle between the two lines.

Answer 1

Hint: Solve the given equation of straight lines to get to seperate equations with other constants then multiply those equations and compare it with the given one to find the value of those constants then find the value of \[\cos \theta \] where \[\theta \] is the angle between the given lines.

Complete step-by-step answer:
We are given the equation of straight lines as \[6{x^2} - 11xy - 10{y^2} - 19y + c = 0\]
If we further try to solve it we will get
 \[\begin{array}{l}
 \Rightarrow 6{x^2} - 11xy - 10{y^2} = 19y - c\\
 \Rightarrow 6{x^2} - 15xy + 4xy - 10{y^2} = 19y - c\\
 \Rightarrow 3x(2x - 5y) + 2y(2x - 5y) = 19y - c\\
 \Rightarrow (3x + 2y)(2x - 5y) = 19y - c
\end{array}\]
Which means that the straight lines will be of the form
\[3x + 2y + {c_1} = 0\& 2x - 5y + {c_2} = 0\]
The pair of lines equation is \[6{x^2} - 11xy - 10{y^2} + x(2{c_1} + 3{c_2}) + (2{c_2} - 5{c_1})y + ({c_1}{c_2}) = 0\]
So if we compare this equation with the equation given to us we will get
\[\begin{array}{l}
 \Rightarrow 2{c_1} + 3{c_2} = 0.............................................(i)\\
 \Rightarrow 2{c_2} - 5{c_1} = - 19.............................................(ii)\\
 \Rightarrow {c_1}{c_2} = c.......................................................(iii)
\end{array}\]
By solving equation (i) and (ii), we will get
\[{c_1} = 3\& {c_2} = - 2\]
Now putting these values in the third equation we will get
\[{c_1}{c_2} = c = - 6\]
Which means that the pair of straight lines is given by the equation \[6{x^2} - 11xy - 10{y^2} - 19y - 6 = 0\]
Now let the angle between the pair of straight lines be \[\theta \]
Then if the equation of lines be
\[\begin{array}{l}
{L_1} \to {a_1}x + {b_1}y + {c_1} = 3x + 2y + 3 = 0\\
{L_2} \to {a_2}x + {b_2}y + {c_2} = 2x - 5y - 2 = 0
\end{array}\]
So the value of \[\cos \theta \] is given by
\[\cos \theta = \dfrac{{\left| {{a_1}{b_1} + {a_2}{b_2}} \right|}}{{\left( {\sqrt {{a_1}^2 + {b_1}^2} } \right)\left( {\sqrt {{a_2}^2 + {b_2}^2} } \right)}}\]
Now if we compare the general equation of lines with the equations we have
We will get
\[\begin{array}{l}
{a_1} = 3\\
{a_2} = 2\\
{b_2} = - 5\\
{b_1} = 2
\end{array}\]
Putting all these values we will get
\[\begin{array}{l}
 \Rightarrow \cos \theta = \dfrac{{\left| {(3)(2) + (2)( - 5)} \right|}}{{\left( {\sqrt {{3^2} + {2^2}} } \right)\left( {\sqrt {{2^2} + {{( - 5)}^2}} } \right)}}\\
 \Rightarrow \cos \theta = \dfrac{4}{{\sqrt {377} }}\\
 \Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{4}{{\sqrt {377} }}} \right)
\end{array}\]

Note: Beware of the signs while doing such bi calculations, especially while multiplying both the equations of straight lines, a simple error in calculation could cause dismissal of the whole answer. Also you need to remember the formula for \[\cos \theta \] .