Question

For the straight lines $4x-3y+10=0$ and $24x-7y+50=0$, the equation of

• A. bisector of the obtuse angle between them is$2x+y=0$
• B. bisector of the acute angle between them is $22x-11y+50=0$
• C. bisector of the angle containing origin is $x+2y=0$
• D. bisector of the angle containing the point (2, 1) is $x+2y=0$

Answer 1

Answer: (B), (C), (D)

The correct options are
B bisector of the angle containing origin is $x+2y=0$
C bisector of the acute angle between them is $22x-11y+50=0$
D bisector of the angle containing the point (2, 1) is $x+2y=0$

$4x-3y+10=0$ and $24x-7y+50=0$

For origin to lie in obtuse traingle,

$a_1\ast a_2+b_1\ast b_2>0$
so,$\left(4\right)\left(24\right)+(-3)(-7)>0$
$\therefore$ origin lies in obtuse angle
Equation of the bisector of the angle in which origin lies is
$\frac{4x-3y+10}{5}=\frac{24x-7y+50}{25}$
i.e. $20x-15y=24x-7y$ i.e. $4x+8y=0$
i.e. $x+2y=0$
Equation of the bisector of the angle in which origin does not lie is
$20x-15y+50=-24x+7y-50$ i.e. $44x-22y+100=0$
i.e. $22x-11y+50=0$
Distance of $(2,1)$ from $x+2y=0$ is $\frac{4}{\sqrt{5}}$
Distance of $(2,1)$ from $22x-11y+50=0$ is $\frac{83}{11\sqrt{5}}$
$\therefore (2,1)$ lies in obtuse angle bisector