Question

The equation of bisector of the acute angles between the lines \[2x + y + 4 = 0\] and \[x + 2y = 1\] is
A) \[x + y + 5 = 0\]
B) \[x + y + 1 = 0\]
C) \[x - 5 = 0\]
D) \[x + 5 = 0\]

Answer 1

Hint: If two lines are of the form \[{a_1}x + {b_1}y + {c_1} = 0\] and \[{a_2}x + {b_2}y + {c_2} = 0\] now the bisector will be given by \[\dfrac{{{a_1}x + {b_1}y + {c_1}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} }} = \pm \dfrac{{{a_2}x + {b_2}y + {c_1}}}{{\sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}\]

Complete step by step answer:
We are given the two lines as \[2x + y + 4 = 0\& x + 2y = 1\]
Which is comparable to \[{a_1}x + {b_1}y + {c_1} = 0\] and \[{a_2}x + {b_2}y + {c_2} = 0\]
So by comparing we can get
\[\begin{array}{l}
{a_1} = 2\\
{b_1} = 1\\
{a_1} = 4\\
{a_2} = 1\\
{b_2} = 2\\
{c_2} = - 1
\end{array}\]
Now we know that the equation of bisectors is given by
\[\dfrac{{{a_1}x + {b_1}y + {c_1}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} }} = \pm \dfrac{{{a_2}x + {b_2}y + {c_1}}}{{\sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}\]
So By applying this we will get
\[\begin{array}{l}
 \Rightarrow \dfrac{{2x + y + 4}}{{\sqrt {{2^2} + {1^2} + {4^2}} }} = \pm \dfrac{{x + 2y - 1}}{{\sqrt {{1^2} + {2^2} + {{( - 1)}^2}} }}\\
 \Rightarrow \dfrac{{2x + y + 4}}{{\sqrt {4 + 1 + 16} }} = \pm \dfrac{{x + 2y - 1}}{{\sqrt {1 + 4 + 1} }}\\
 \Rightarrow \dfrac{{2x + y + 4}}{{\sqrt {21} }} = \pm \dfrac{{x + 2y - 1}}{{\sqrt 6 }}
\end{array}\]
Now let us check the value of
\[{a_1}{a_2} + {b_1}{b_2}\]
Which is indeed
\[\begin{array}{l}
 = 2 \times ( - 1) + 1 \times ( - 2)\\
 = - 2 + ( - 2)\\
 = - 2 - 2\\
 = - 4
\end{array}\]
Now we can clearly see that \[{a_1}{a_2} + {b_1}{b_2} < 0\]
Which means that we will be taking the negative sign, i.e.,
\[\dfrac{{2x + y + 4}}{{\sqrt {21} }} = - \dfrac{{x + 2y - 1}}{{\sqrt 6 }}\]
Which after solving we will get it as \[x + y + 1 = 0\] and option B is the correct option here.

Note:
If the value of \[{a_1}{a_2} + {b_1}{b_2} > 0\] then we could have took the value \[\dfrac{{{a_1}x + {b_1}y + {c_1}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} }} = \dfrac{{{a_2}x + {b_2}y + {c_1}}}{{\sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}\] .