The angle between the lines represented by the $\lambda {x^2} + {\left( {1 - \lambda } \right)^2}xy - \lambda {y^2} = 0$ is
A. ${30^ \circ }$
B. ${45^ \circ }$
C. ${60^ \circ }$
D. ${90^ \circ }$

Hint: We have to find the angle between the lines for that first we will compare the given equation $\lambda {x^2} + {\left( {1 - \lambda } \right)^2}xy - \lambda {y^2} = 0$ with general equation $a{x^2} + 2hxy + b{y^2} = 0$ and find the value of $a$,$b$ and $h$. Then all these values in the formula for the angle between the curves to find the angle.

Formula Used:
$\tan \alpha = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$

Complete step by step solution: Given, equation is $\lambda {x^2} + {\left( {1 - \lambda } \right)^2}xy - \lambda {y^2} = 0$
The general equation is $a{x^2} + 2hxy + b{y^2} = 0$
On comparing, we will get
$a = \lambda $, $h = {(1 - \lambda )^2}$ and $b = - \lambda $
We know the angle between the two lines is
$\tan \alpha = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
\[\tan \alpha = \left| {\dfrac{{2\sqrt {\left( {\dfrac{{{{\left( {1 - \lambda } \right)}^2}}}{2}} \right) - (\lambda )( - \lambda )} }}{{\lambda + \left( { - \lambda } \right)}}} \right|\]
After solving, we get
\[\tan \alpha = \left| {\dfrac{{2\sqrt {\left( {\dfrac{{{{\left( {1 - \lambda } \right)}^2}}}{2}} \right) - (\lambda )( - \lambda )} }}{0}} \right|\]
Taking ${\tan ^{ - 1}}$ on both sides
\[{\tan ^{ - 1}}\left( {\tan \alpha } \right) = {\tan ^{ - 1}}\left( {\left| {\dfrac{{2\sqrt {\left( {\dfrac{{{{\left( {1 - \lambda } \right)}^2}}}{2}} \right) - (\lambda )( - \lambda )} }}{0}} \right|} \right)\]
$\alpha = {90^ \circ }$
Hence, the angle between the lines is ${90^ \circ }$

Therefore, option D is correct

Note: Students should solve the question correctly to avoid any mistakes. They should know that $\tan {90^ \circ }$ is undefined if they do not know about that they might get confused while solving for $\alpha $.