Question

Find equation of the line bisecting the angles between the line $Ax+By+C=0$ and the line passing through the point $(h,k)$ and perpendicular to this line.

Answer 1

Hint:In this problem we will first find the equation of the line passing through $(h,k)$ and perpendicular to the line $Ax+By+C=0$. The equation of a line perpendicular to a given line $Ax+By+C=0$ is $Bx-Ay+\lambda =0......(i)$ where $\lambda $ is a constant. The value of $\lambda $ can be found out by some given condition.After finding the equation of this line, we can find the equations of the bisectors of the angle between two lines obtained above. This can be obtained using the formula given below:
If ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ are the two lines, then the equations of the bisectors of the lines are given by,
$\dfrac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}}=\pm \dfrac{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}}{\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}}}........(ii)$
We will be using these two formulas in this question to obtain the answers.

Complete step-by-step answer:
It is given that there are two lines, $Ax+By+C=0$ and a line passing through $(h,k)$ and perpendicular to $Ax+By+C=0$. To find this line we will use equation (i).
The equation of a line perpendicular to $Ax+By+C=0$ according to equation (i) is,
$\Rightarrow Bx-Ay+\lambda =0$
Now, we have to find $\lambda $. In the question it is given that the line passes through $(h,k)$.
Now, substituting the values $(h,k)$ in the above equation,
$\begin{align}
  & \Rightarrow Bh-Ak+\lambda =0......(iii) \\
 & \Rightarrow \lambda =-Bh+Ak......(iv) \\
\end{align}$
Substituting equation (iv) in equation (i) we get,
$\Rightarrow Bx-Ay+(-Bh+Ak)=0........(v)$
Simplifying the above equation, we can also write it as,
$\begin{align}
  & \Rightarrow Bx-Bh-Ay+Ak=0 \\
 & \Rightarrow B(x-h)-A(y-k)=0........(vi) \\
\end{align}$
Thus, equation (vi) is the required equation of line passing through $(h,k)$ and perpendicular to $Ax+By+C=0$.
Now, we have to find the equations of the lines bisecting the angles between the lines$Ax+By+C=0$ and $Bx-Ay+(-Bh+Ak)=0$.
Comparing with equation (ii) we can write the equations of the lines bisecting these lines as,
$\Rightarrow \dfrac{Ax+By+C}{\sqrt{{{A}^{2}}+{{B}^{2}}}}=\pm \dfrac{Bx+(-Ay)+(-Bh+Ak)}{\sqrt{{{B}^{2}}+{{A}^{2}}}}.........(vii)$
Now, simplifying the above equation we get,
$\begin{align}
  & \Rightarrow \dfrac{Ax+By+C}{{\sqrt{{{A}^{2}}+{{B}^{2}}}}}=\pm \dfrac{Bx+(-A)y+(-Bh+Ak)}{{\sqrt{{{B}^{2}}+{{A}^{2}}}}} \\
 & \Rightarrow Ax+By+C=\pm (Bx-Ay+(-Bh+Ak)) \\
 & \Rightarrow Ax+By+C=\pm (B(x-h)-A(y-k)).......(viii) \\
\end{align}$
$\therefore $ Equation (viii) represents the equations of the lines bisecting the angles between the lines$Ax+By+C=0$ and $Bx-Ay+(-Bh+Ak)=0$.
Hence, the answer is $Ax+By+C=\pm (B(x-h)-A(y-k))$.

Note: The main idea in this problem is that the student should know how to formulate the equation of a line perpendicular to another line passing through a point (x, y) and also how to formulate the equations of the bisectors of the angle between two lines.The equation of a line perpendicular to a given line $Ax+By+C=0$ is $Bx-Ay+\lambda =0......(i)$ In equation (i) $\lambda $ should be evaluated with care.Also we have to remember the equations of the bisectors of the lines which is given by
$\dfrac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}}=\pm \dfrac{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}}{\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}}}$.