Question

The equation of the bisector of the angle between two lines 3x -4y+12 = 0 and 12x-5y+7 = 0 which contains the points (-1, 4) is :

• A. 21x+27y-121 = 0
• B. 21x-27y-121 = 0
• C. 21x+27y+191 = 0
• D. $\frac{-3x+4y-12}{5}=\frac{12x-5y+7}{13}$

Answer 1

The correct option is A 21x+27y-121 = 0

$\begin{array}{cc}{L}_1=3x-4y+12& =0\\ L_2=& 12x-5y+7=0\\ & \text{points}\left(p\right)=(-1,4)\\ & \text{Putting point}P\text{in}{L}_1\text{and}{L}_2\\ & \left(L_1\right)\times L_2\\ \Rightarrow & (-3-4\times 4+12)(-12-5\times 4+7)\\ \Rightarrow & (-7)(-25)>0\\ & \text{(positive)}\end{array}$

$\frac{a_1{x}_1+b{y}_1+c_1}{\sqrt{{\left(a_1\right)}^2+(b{)}^2}}=\frac{a_2{x}_1+b_2{x}_1+c_2}{\sqrt{{\left(a_2\right)}^2+{\left(b_2\right)}^2}}$

$\frac{a_{1}x+b_{1}y+c_1}{\sqrt{{\left(a_1\right)}^2+{\left(b_1\right)}^2}}=\frac{a_{2}x+b_{2}y+c_2}{\sqrt{{\left(a_2\right)}^2+{\left(b_2\right)}^2}}$

$\Rightarrow \frac{3x-4y+12}{\sqrt{(3{)}^2+(4{)}^2}}=\frac{+2x-5y+7}{\sqrt{(12{)}^2+(5{)}^2}}$

$\Rightarrow \frac{3x-4y+12}{\sqrt{25}}=\frac{12x-5y+7}{\sqrt{169}}$

$\Rightarrow 13(3x-4y+12)=5(12x-5y+7)$

$\Rightarrow 39x-52y+156=60x-25y+35$

$\Rightarrow 60x-39x-25y+52y+35-156=0$

$\Rightarrow 21x+27y-121=0$