Question

Find the value of $\lambda$ in $\lambda x^2-10xy+12y^2+5x-16y-3=0$ which represents a pair of straight lines .
A. $1$
B. $-1$
C. $2$
D. $-2$

Answer 1

Hint: In the given question we have to find $\lambda$ in the equation for which we will use the general form of a pair of straight lines to equate the variables .

Formula Used:
We have to use the general equation of Pair of Straight line which is $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0.$

Complete step by step solution:
Given,
$\lambda x^2-10xy+12y^2+5x-16y-3=0$ -------------- (i)
For finding out the value of $\lambda$ from the equation$\lambda x^2-10xy+12y^2+5x-16y-3=0$we have to equate the equation with the general equation of Pair of Straight line which is $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0.$
And while using the general equation we will be following the rule where the Determinant will be equal to 0.
$\therefore \mathrm{\Delta }=0$
$\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=0$
Now we have to equate the equation (i) with the general equation to find out the variables.
After Equating we will get
$a=\lambda ,b=12,h=-5,g={\displaystyle \frac{5}{2}},f=-8,c=-3$
Putting the values of the variables in the determinant.
$\left|\begin{array}{ccc}\lambda & -5& {\displaystyle \frac{5}{2}}\\ -5& 12& -8\\ {\displaystyle \frac{5}{2}}& -8& -3\end{array}\right|=0$
Solving the determinant:
$\lambda \left(-36-64\right)+5\left(15+20\right)+\left({\displaystyle \frac{5}{2}}\right)\left(40-30\right)=0$
$-100\lambda +175+25=0$
$-100\lambda =-200$
Hence,
$\lambda =2$

Option ‘C’ is correct

Note: Solve this question using the determinant method. Generally we go wrong in finding out the variables and determinants using the general equation and while solving this question the general formula should be remembered that is $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0.$