Question

The equation of the line which bisects the obtuse angle between the lines $x-2y+4=0$ and $4x+3y+2=0$, is

• A. $(4-\sqrt{5})x-(3-2\sqrt{5})y+(2-4\sqrt{5})=0$
• B. $(4+\sqrt{5})x+(3-2\sqrt{5})y+(2+4\sqrt{5})=0$
• C. $(4+\sqrt{5})x-(3-2\sqrt{5})y+(2-4\sqrt{5})=0$
• D. $Noneofthese$

Answer 1

The correct option is B $(4+\sqrt{5})x+(3-2\sqrt{5})y+(2+4\sqrt{5})=0$

$L_1:x-2y+4=0$

$L_2:4x+3y+2=0$

$\text{Now,}{a}_1{a}_2+b_1{b}_2$

$=4-6$

$=-2<0$

$\text{So, obtuse}\angle \text{bisector with}-\text{ve sign.}$

$\Rightarrow \frac{x-2y+4}{\sqrt{1+4}}=-\frac{(4x+3y+2)}{\sqrt{16+9}}$

$\Rightarrow \frac{x-2y+4}{\sqrt{5}}=-\frac{4x-3y-2}{5}$

$\Rightarrow 5x-10y+20=-4\sqrt{5}x-3\sqrt{5}y-2\sqrt{5}$

$\Rightarrow \sqrt{5}x-2\sqrt{5}y+4\sqrt{5}=-4x-3y-2$

$\Rightarrow (4x+\sqrt{5}x)+(3y-2\sqrt{5}y)+2+4\sqrt{5}=0$

$\Rightarrow (4+\sqrt{5})x+(3-2\sqrt{5})y+(2+4\sqrt{5})=0$