Question

$L_1$ = x-2y+11 = 0 and $L_2$ = 3x+6y+5 = 0

• A. Equation of acute angle bisector of is y = .
• B. Equation of acute angle bisector of is y = - .
• C. Equation of obtuse angle bisector of is x = .
• D. Equation of obtuse angle bisector of is x = - .

Answer 1

The correct option is A

Equation of acute angle bisector of is y = .

We know how to find the bisectors. Once we find them , we have to distinguish between obtuse and acute angle bisectors.

If $\theta$ be the angle between one of the line and one of the bisectors, find tan $\theta$ to distinguish between obtuse and acute

angle bisectors. If tan $\theta$ < 1, the bisector is acute angle bisector, otherwise obtuse angle ( If tan $\theta$ = 1 , then we can't distinguish )

1 ) Finding the bisectors

$\frac{x-2y+11}{\sqrt{1^2+2^2}}=\pm \frac{3x+6y+5}{\sqrt{3^2+6^2}}$
$\Rightarrow \frac{x-2+11}{\sqrt{5}}=\pm \frac{3x+6y+5}{\sqrt{3\sqrt{5}}}$

$\Rightarrow$ 3x - 6y + 33 = 3x + 6y + 5 or

$\Rightarrow$ 3x - 6y + 33 = -3x - 6y - 5

$\Rightarrow$ 12y = 28 or 6x = -38

$\Rightarrow$ 3y = 7 or 3x = -19

$\Rightarrow$ 3x - 6y + 33 = - 3x - 6y - 5

$\Rightarrow$ 12y = 28 or 6x = -38

$\Rightarrow$ 3y = 7 or 3x = -19

2) Distinguish between obtuse and acute angle bisectors

y = $\frac{7}{3}$ is a line parallel to x - axis

so , its slope is zero

Slope of x - 2y + 11 = $\frac{1}{2}$

if $\theta$ is the angle between y = $\frac{7}{3}$ and x - 2y + 11, then

| $tan\theta |=\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{\frac{1}{2}-0}{1+0\ast \frac{1}{2}}\right|=\frac{1}{2}$

$|tan\theta |$ < $1\Rightarrow$ $y=\frac{7}{3}$ is the acute angle bisector.

$\Rightarrow$ x = $\frac{-19}{3}$ is the obtuse angle bisector.