Question

What is the equation of the bisector of the acute angle between the lines $3x-4y+7=0$ and $12x+5y-2=0$?

• A. $21x+77y-101=0$
• B. $11x-3y+9=0$
• C. $31x+77y+101=0$
• D. $11x-3y-9=0$

Answer 1

The correct option is B

$11x-3y+9=0$

Explanation of correct answer :

Finding the equation of the bisector of the acute angle between the lines :

Given, equation of lines $3x-4y+7=0\dots \left(1\right)$ and

$12x+5y-2=0\dots \left(2\right)$

In equation $\left(2\right)$, multiply it with $-1$.

$\Rightarrow -12x-5y+2=0$

Here $a_1{a}_2+b_1{b}_2=3\left(-12\right)+4\times 5$

$\Rightarrow a_1{a}_2+b_1{b}_2=-16<0$

For acute angle bisector, we take the positive sign.

$$\begin{gathered} \frac{a_{1}x+b_{1}y+c_1}{\sqrt{{a_1}^2+{b_1}^2}}=\frac{a_{2}x+b_{2}y+c_2}{\sqrt{{a_2}^2+{b_2}^2}} \\ \Rightarrow \frac{3x-4y+7}{\sqrt{3^2+4^2}}=\frac{-12x-5y+2}{\sqrt{{12}^2+5^2}} \\ \Rightarrow \frac{3x-4y+7}{5}=\frac{-12x-5y+2}{13} \\ \Rightarrow 39x-52y+91=-60x-25y+10 \\ \Rightarrow 99x-27y+81=0 \\ \Rightarrow 11x-3y+9=0 \end{gathered}$$

Hence, the equation of the bisector of the acute angle is $11x-3y+9=0$

Thus, the correct answer is Option(B).