Question

The value of $\lambda$, for which the equation $6x^2+11xy-10y^2+x+3+y\lambda =0$ represents a pair of straight lines, is

• A. $-15$
• B. $0$
• C. $2$
• D. None of these

Answer 1

Answer: (A)

$-15$

Given, equation will represent a pair of straight lines, if
$△=abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$
$h^2-ab=0$
Here, $a=6,h=\frac{1}{2},b=-10$,
$g=\frac{1}{2},f=\frac{31}{2},c=\lambda$
$\therefore △=0$
$\Rightarrow \lambda =\frac{-5415}{361}=-15$
Also, $h^2-ab=\frac{121}{4}+60=\frac{361}{4}>0$
Hence, $\lambda =-15$ is valid.