The angle between the lines ${x^2} - xy - 6{y^2} - 7x + 3y - 18 = 0$ is
A. ${60^ \circ }$
B. ${45^ \circ }$
C. ${30^ \circ }$
D. ${90^ \circ }$

Hint: In order to solve this type of question, first we need to write the given equation and the general equation. Then, we have to compare both the equations to get the values of a, h and b. Next, we will use the formula for finding the angle between the lines and substitute the values in it to get the correct answer.

Formula used:
$a{x^2} + b{y^2} + 2hxy + 2gx + 2fy + c = 0$
$\tan \theta = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$

Complete step by step solution:
We are given the following pair of lines,
${x^2} - xy - 6{y^2} - 7x + 3y - 18 = 0$ ………………….equation$\left( 1 \right)$
We know that the general equation is,
$a{x^2} + b{y^2} + 2hxy + 2gx + 2fy + c = 0$ ………………….equation$\left( 2 \right)$
On comparing equation $\left( 1 \right)$ and $\left( 2 \right)$ we get,
$a = 1,\;h = \dfrac{{ - 1}}{2},\;b = - 6$
Finding the angle between the lines,
$\tan \theta = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
Substituting the values of a, h and b,
$\tan \theta = \left| {\dfrac{{2\sqrt {{{\left( {\dfrac{{ - 1}}{2}} \right)}^2} - 1 \times \left( { - 6} \right)} }}{{1 + \left( { - 6} \right)}}} \right|$
$\tan \theta = \left| {\dfrac{{2\sqrt {\dfrac{1}{4} + 6} }}{{ - 5}}} \right|$
Solving it,
$\tan \theta = \left| {\dfrac{{2\sqrt {\dfrac{{25}}{4}} }}{{ - 5}}} \right|$
$\tan \theta = 1$
Solving for $\theta ,$ we get,
$\theta = {45^ \circ }$
$\therefore $ The correct option is B.

Note: Do not make any calculation mistakes while substituting and solving the equation to get the correct angle. Solve this type of question by being very sure while writing the general equation and comparing the values of a, h and b. We should not bother about the values of f and g while calculating the angle between the pair of lines.