Question

The angle between the lines in $x^2-xy-6y^2-7x+31y-18=0$ is

• A. $90°$
• B. $60°$
• C. $45°$
• D. $30°$

Answer 1

The correct option is C

$45°$

Find the angle between the lines

Consider the given equation, $x^2-xy-6y^2-7x+31y-18=0......\left(1\right)$

We know that standard equation is $a{x}^2+2hxy+b{y}^2+2gx+2hy+c=0......\left(2\right)$

On comparing equation (1) and (2), we get;

$a=1,b=-6,h=-\frac{1}{2}$

We know that angle between the lines is given by;

$\tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$

Substitute the value $a=1,b=-6,h=-\frac{1}{2}$ in the above equation, we get;

$$\begin{gathered} \Rightarrow \tan \theta =\left|\frac{2\sqrt{{\left(-{\displaystyle \frac{1}{2}}\right)}^2-\left(1\right)\left(-6\right)}}{\left(1\right)+\left(-6\right)}\right| \\ \Rightarrow \tan \theta =\left|\frac{2\sqrt{{\displaystyle \frac{1}{4}}+6}}{-5}\right| \\ \Rightarrow \tan \theta =\left|\frac{2\sqrt{{\displaystyle \frac{25}{4}}}}{-5}\right| \\ \Rightarrow \tan \theta =\left|\frac{2{\displaystyle \times \frac{5}{2}}}{-5}\right| \\ \Rightarrow \tan \theta =\frac{5}{5} \\ \Rightarrow \tan \theta =1 \\ \Rightarrow \theta ={\tan }^{-1}1 \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{4} \\ \Rightarrow \theta =45° \end{gathered}$$

Hence, the correct answer is Option C.