Question

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

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

Answer 1

The correct option is A

$45°$

Explanation for the correct option:

Finding the angle between the lines,

Given the equation of circle is

$x^2-xy-6y^2-7x+31y-18=0\left(1\right)$

We know that the general equation of circle is,

$a{x}^2+2hxy+b{y}^2+2gx+2fxy+c=0\left(2\right)$

Now, comparing the equations $\left(1\right)$ and $\left(2\right)$ we get,

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

Finding the angle between the two lines of equation using the formula,

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

Substitute the values of $a=1,b=-6,h=\frac{-1}{2}$ to find the angle
$\begin{array}{rcl}\tan \theta & =& \left|\frac{2\sqrt{{\left(-{\displaystyle \frac{1}{2}}\right)}^2-\left(1\right)(-6)}}{1+(-6)}\right|\\ & =& \left|\frac{2\sqrt{{\displaystyle \frac{1}{4}}+6}}{1-6}\right|\\ & =& \left|\frac{2\sqrt{{\displaystyle \frac{(1+24)}{4}}}}{-5}\right|\\ & =& \left|\frac{2\sqrt{{\displaystyle \frac{25}{4}}}}{-5}\right|\\ & =& \left|\frac{2\left({\displaystyle \frac{5}{2}}\right)}{-5}\right|\\ & =& \left|\frac{2}{-5}\times \frac{5}{2}\right|\\ & =& |-1|\\ \therefore \tan \theta & =& 1\text{or}-1\end{array}$
By taking only the positive value of $\tan \theta$, we get

$\begin{array}{rcl}\tan \theta & =& 1\\ \theta & =& {\tan }^{-1}\left(1\right)\\ \theta & =& 45°\end{array}$

Therefore, the angle between the two lines is $45°$.

Hence, option (A) is the correct answer.