Question

For the straight lines $4x+3y-6=0$ and $5x+12y+9=0$ the equation of the bisector of the acute angle between between them = $7x+9y-3=0$

• A. True
• B. False

Answer 1

The correct option is A True

To find the bisector of the angle between the lines which contains the origin, we first write down the equations of the given lines in such a form that the constant terms in the equations of the lines are positive. The equations of the given lines are

$4x+3y-6=0\Rightarrow -4x-3y+6=0.....\left(i\right)$

$5x+12y+9=0....\left(ii\right)$

Now the equation of the bisector of the angle between the lines which contains the origin is the bisector corresponding to the positive symbol i.e.,

$\frac{-4x-3y+6}{\sqrt{{(-4)}^2+{(-3)}^2}}=\frac{5x+12y+9}{\sqrt{5^2+{12}^2}}\Rightarrow \frac{-4x-3y+6}{\sqrt{16+9}}=\frac{5x+12y+9}{\sqrt{25+144}}\Rightarrow \frac{-4x-3y+6}{\sqrt{25}}=\frac{5x+12y+9}{\sqrt{169}}\Rightarrow \frac{-4x-3y+6}{5}=\frac{5x+12y+9}{13}$

$\Rightarrow 13(-4x-3y+6)=5(5x+12y+9)\Rightarrow -52x-39y+78=25x+60y+45\Rightarrow 25x+52x+60y+39y+45-78=0\Rightarrow 77x+99y-33=0\Rightarrow 11(7x+9y-3)=0\Rightarrow 7x+9y-3=0$

From equations (i) and (ii), we have $a_1{a}_2+b_1{b}_2=-20–36=-56<0$.

Hence, the origin is situated in an acute angle region and the bisector of this angle is $7x+9y–3=0$.