Question

If the angle between the two lines represented by $2x^2+5xy+3y^2+6x+7y+4=0$ is ${\tan }^{-1}\left(m\right),$ then the value(s) of $m$ is/are

• A. $\frac{1}{5}$
• B. $-5$
• C. $-\frac{1}{5}$
• D. $5$

Answer 1

Answer: (A), (C)

$-\frac{1}{5}$

The angle between the lines represented by $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ is given by ${\tan }^{-1}(\pm \frac{2\sqrt{h^2-ab}}{a+b})$
So, the angle between the lines represented by $2x^2+5xy+3y^2+6x+7y+4=0$ is
${\tan }^{-1}m={\tan }^{-1}(\pm \frac{2\sqrt{h^2-ab}}{a+b})$
Here, $a=2,b=3,h=\frac{5}{2}$
$\Rightarrow {\tan }^{-1}m={\tan }^{-1}(\pm \frac{2\sqrt{\frac{25}{4}-6}}{2+3})$
$\Rightarrow {\tan }^{-1}m={\tan }^{-1}(\pm \frac{1}{5})$
$\Rightarrow m=\pm \frac{1}{5}$

Alternate Solution :
$2x^2+5xy+3y^2+6x+7y+4=0\Rightarrow (x+y)(2x+3y)+(2x+3y)+4(x+y)+4=0\Rightarrow (2x+3y)((x+y)+1)+4(x+y+1)=0\Rightarrow (x+y+1)(2x+3y+4)=0$
So lines will be $x+y+1=0;2x+3y+4=0$
So angle between the lines will be
${\tan }^{-1}m={\tan }^{-1}(\pm \frac{-1+\frac{2}{3}}{1+\frac{2}{3}})\Rightarrow m=\pm \frac{1}{5}$