Question

Find the equation of the angle bisectors of the lines 3x-4y+8 = 0 and 8x+6y-9 = 0

• A. 2x+14y-25 = 0
• B. 2x+4y+25 = 0
• C. 14x-2y+7 = 0
• D. 14x-2y-7 = 0

Answer 1

Answer: (A), (C)

The correct options are
A

2x+14y-25 = 0

C

14x-2y+7 = 0

Equations of angle bisectors of $a_{1}x+b_{1}y+c_1=0and{a}_{2}x+b_{2}y+c_2=0$ is given by

\\frac{a1x+b1y+c1}{\sqrt{a^2_1+b^2_1}}= \pm \frac{a_2x+b_2y+c_2}{\sqrt{a^2_2+b^2_2}}\)

We can use this formula to find the equations of bisectors. We will first discuss the method to find it without formula.
We can find the point of intersection of the two lines given. Bisectors also pass through this point. Now, if we know the slope of those two bisectors, we can find their equations.

We also know that the two bisectors are perpendicular to each other. So, if we know the slope of one of the bisectors, we can find the slope of the other easily.

To find the slope of $L_1$ , we can find the angle between L and $L_1$. We can use this to find the slope of $L_1$.

We can see that this method is lenghty. So we will always go for formula

$\Rightarrow$ equations are

$\frac{3x-4y+8}{\sqrt{3^2+4^2}}=\pm \frac{8x+6y-9}{\sqrt{8^2+6^2}}$
$\Rightarrow \frac{3x-4y+8}{5}=\pm \frac{8x+6y-9}{10}$
$\Rightarrow$ 3x-4y+8 = ± $\frac{(8x+6y-9)}{2}$

$\Rightarrow$ 2(3x-4y+8) = 8x + 6y - 9

Or

2(3x-4y+8) = - (82 + 6y - 9)

$\Rightarrow$ 6x - 8y + 16 = 8x + 6y - 9

Or

6x - 8y + 16 = -8x - 6y + 9

$\Rightarrow$ 2x+14y = 25 or 14x - 2y + 7 = 0