Question

The straight lines represented by $x^2+mxy-2y^2+3y-1=0$ meet at

• A. $(-\frac{1}{3},\frac{2}{3})$
• B. $(-\frac{1}{3},-\frac{2}{3})$
• C. $(\frac{1}{3},\frac{2}{3})$
• D. one of these

Answer 1

Answer: (A), (C)

$(-\frac{1}{3},\frac{2}{3})$
$(\frac{1}{3},\frac{2}{3})$

$x^2+mxy-2y^2+3y-1=0$
Represent pair of lines when $\Delta =0\Rightarrow$$m=\pm 1$
For $m=1$
$x^2+xy-2y^2+3y-1=0\Rightarrow (x-y+1)(x+2y-1)=0$
Hence intersection point is $(-\frac{1}{3},\frac{2}{3})$
For $m=-1$
$x^2-xy-2y^2+3y-1=0\Rightarrow (x-2y+1)(x+y-1)=0$
Hence intersection point is $(\frac{1}{3},\frac{2}{3})$